1.
Four boys and five girls, all of differing heights, stood in line for a palm reading. If all the boys stand next to each other in line, how how many different linear arrangements exist for the nine people? Solution: 4!*6!
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The four boys, standing next to each other, can be arranged in P(4,4)=4! ways. Now treat them as one unit, because they the y cannot be separated. This gives us six units to permute for possible linear arrangements, carried out in P(6,6)=6! ways. 2.
A local fast-food outlet offered a variety of meal combinations. Every meal combination included a sandwich, an order of French fries, and a soft drink. Suppose there are 6 different sandwiches, 3 different sizes for French fries orders, and 8 different soft drinks to choose from. (a) How many meal combo orders must be placed to assure that at least one meal combo is ordered twice? (4 points) Solution: (6*3*8)+1=144+1=1 (6*3*8)+1=144+1=145 45 Use the multiplication principle to determine there are 6*3*8=144 different combo meals. The worst case scenario is that the first 144 orders are each for different combo meals. By the pigeonhole principle, the 145th order, therefore, must assure us that at least one combo meal is ordered twice. The fast-food outlet also offers breakfast items. The breakfast sandwiches include five different bagel sandwiches, seven s even different biscuit sandwiches, and four different English muffin sandwiches. sandwiches. (b) Suzzie Softknuckle comes to the fast-food outlet for a breakfast sandwich. How many different breakfast sandwiches does she have to choose from? (3 points) Solution: 16 choices The sandwich types are disjoint from one another, so use the addition principle to determine there are 5+7+4 different sandwiches. The Merchanteer County All-Stars softball team stopped at the fastfood outlet. Each of the 21 team members purchased either a double cheeseburger or a hot ham-and-cheese sandwich. No team member ordered more than one sandwich. (c) Freddie the Fry Cook kept track of the sandwich purchases in the exact order they were made. How many different orderings were possible if k team members purchased a hot ham-and-cheese
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sandwich? (3 points) Solution: C(21,k)=C(21,21-k) In the list of 21 sandwich orders, select k of them for hot ham-andcheese. This can be done in C(21,k) ways. 3.
Five brothers each have the same set of seven hats, distinguished only by color. Each has a white hat, a black hat, an orange hat, a green hat, a yellow hat, a maroon hat, and a red hat. (a) If each brother chooses a hat to wear, how many 5-hat sets could they be seen wearing? (3 points) Solution: 7^5 Each brother chooses one of seven hats. Each brother's choice is independent of the others, so by the multiplication principle, there are 7*7*7*7*7 different 5-hat sets. (b) How many ways are there for each of the brothers to all choose hats of different colors? (3 points)
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Solution: P(7,5) The first brother has 7 colors to choose from, the next has 6, and so on, down to the last brother who has 3 colors to choose from. (c) At a recent family famil y reunion, four of the brothers were seen wearing the same color hat while the fifth fi fth brother wore a hat of a different color. In how many ways could this have occurred? (4 points) Solution: P(7,2)*C(5,1)=7*6*5 Choose one of the five brothers to wear the different-colored hat. This can be done in C(5,1)=5 ways. The set of four brothers choose one of seven colors and the fifth chooses from the remaining six colors. 4.
4. Consider the letters in the word EXCESSIVENESS. (a) How many unique arrangements are there for the letters in this word? (2 points) Solution: 13!/(4!1!1!4!1!1!1!)=13!/(4!4!) We permute the 13 letters in the word, carried out in 13! ways. We then divide by the denominator shown here to account for the duplicate letters that occur in the word.
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sandwich? (3 points) Solution: C(21,k)=C(21,21-k) In the list of 21 sandwich orders, select k of them for hot ham-andcheese. This can be done in C(21,k) ways. 3.
Five brothers each have the same set of seven hats, distinguished only by color. Each has a white hat, a black hat, an orange hat, a green hat, a yellow hat, a maroon hat, and a red hat. (a) If each brother chooses a hat to wear, how many 5-hat sets could they be seen wearing? (3 points) Solution: 7^5 Each brother chooses one of seven hats. Each brother's choice is independent of the others, so by the multiplication principle, there are 7*7*7*7*7 different 5-hat sets. (b) How many ways are there for each of the brothers to all choose hats of different colors? (3 points)
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Solution: P(7,5) The first brother has 7 colors to choose from, the next has 6, and so on, down to the last brother who has 3 colors to choose from. (c) At a recent family famil y reunion, four of the brothers were seen wearing the same color hat while the fifth fi fth brother wore a hat of a different color. In how many ways could this have occurred? (4 points) Solution: P(7,2)*C(5,1)=7*6*5 Choose one of the five brothers to wear the different-colored hat. This can be done in C(5,1)=5 ways. The set of four brothers choose one of seven colors and the fifth chooses from the remaining six colors. 4.
4. Consider the letters in the word EXCESSIVENESS. (a) How many unique arrangements are there for the letters in this word? (2 points) Solution: 13!/(4!1!1!4!1!1!1!)=13!/(4!4!) We permute the 13 letters in the word, carried out in 13! ways. We then divide by the denominator shown here to account for the duplicate letters that occur in the word.
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(b) How many arrangements exist if the three-letter sequence EXC must be kept together in the order shown? (2 points) Solution: 11!/(1!4!4!1!1!1!) Treat EXC as a unit, resulting in 11 objects to permute. We now carry out the same process as in (a) above. (c) How many arrangements exist if each must begin and end with a consonant? (3 points) Solution: [P(8,2)*P(11,11)]/(4!*4!)=(8 [P(8,2)*P(11,11)]/(4!*4!)=(8*7*11!)/(4!*4!) *7*11!)/(4!*4!) Within the 13 places for letters, place the first and last to assure they are consonants. There are 8 consonants to choose from for the first and 7 for the last. This is P(8,2)=8*7. Now place the remaining 11 letters. This can be done in P(11,11)=11! ways. Because each of the t he P(8,2) possible first-last consonant arrangements can be matched with the P(11,11) ways to place the remaining letters, we multiple the two results. Finally, we account for the duplication of letters, let ters, requiring us to divide out those equivalent groups. We divide by (4!*4!) as we did in (a) above. This results in [P(8,2)*P(11,11)]/(4!*4!)=(8*7*11!)/(4!*4!) [P(8,2)*P(11,11)]/(4!*4!)=(8*7* 11!)/(4!*4!) ways to rearrange rearr ange the letters with a consonant in the first fi rst place and a consonant in the last place. (d) How many 3-letter sets can be created using only the unique letters in the word? (3 points) Solution: C(7,3) There are 7 unique letters in the word and we select 3 of them. 5.
At Bunion College, a small undergraduate institution in the midwest, every student must select a password to use to enter the college computer network. In creating a password, the following restrictions must all be met. •
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The password must contain only digits or letters of the alphabet. The password must be four or five characters in length. The password must begin with either a B or a C.
How many different passwords are possible under these restrictions? Solution: 2*36^3 + 2*36^4
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Consider two cases, one for the 4-character password and one for the 5-character password. Case I: For a 4-character password, there are 2 choices for the first position (B or C) and 36 choices for each of the next three positions, resulting in 2*36^3 different 4-character passwords. Case I: For a 5-character password, there are again 2 choices for the first position (B or C) and 36 choices for each of the next four positions, resulting in 2*36^4 different 5-character passwords. Because the cases are disjoint&emdash;we cannot have a password that is both 4 and 5 characters long&emdash;we can add the results of the two cases. 6.
Francisco approached his combinatorics instructor and showed her the following claims: (i) 5! + 5! = 10! (ii) 2! &endash; 1! = 1! (iii) 5! = 5*4! (a) State whether each claim is true or false. Show arithmetic work to support your response. (3 points) Solution (i) 5! + 5! = 10! False: 5!=120, so 5!+5!=240 but 10!=3,628,800. (ii) 2! - 1! = 1! True: 2!=2 and 1!=1, so 2-1=1 is true. (iii) 5! = 5*4! True: 5!=5*4*3*2*1, but 4*3*2*1=4!, so 5!=5*4!. (b) For any of the three claims that are true, write a generalization of Francisco's claim. (3 points) Generalization: n!-(n-1)!=(n-1)! Or, alternatively, n!1!=1! (iii) 5! = 5*4! Generalization: n!=n*(n-1)! (ii) 2! - 1! = 1!
(c) For each generalization you wrote for (b), determine whether it is true or false. If it is false, provide a counterexample to show that; if it is true, justify that the result holds in general. (4 points) (ii-a) n!-(n-1)!=(nFalse: Counterexample: 6!-5! is not equal to 5! 1)! (ii-b) n!-1!=1! False: Counterexample: 10!-1! is not equal to 1! True: By the definition of factorials this must (iii) n!=n*(n-1)! always hold.
1.
Respond to each of these questions. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) on this page. (2 points each) (a) What value r satisfies the equation C(42,19) = C(42,r)?
Solution: r = 23 (b) How many different arrangements exist for the letters in the word antivivisectionist ?
Solution: 18!/(2!3!5!2!2!) In the expansion of (v+w+x+y+x)^20 , state: (c) the number of uncollected terms,
Solution: 5^20 (d) the number of collected terms,
Solution: C(24,4) (e) the coefficient T in the collected term Tv^13wy^2z^4.
Solution: 20!/(13!1!0!2!4!) 2.
Solve each of the following counting problems. (2 points each) (a) You are ordering a 5-course dinner at a fancy restaurant. For each course, you have 7 choices. How many different dinners can you order?
Solution: 7^5 (b) Ten people arrive for a casting call and four are chosen, to play the roles of Annie, Bud, Cathy, and Dianne, respectively. In how many ways can such a cast be created from the 10 people who arrived?
Solution: P(10,4) (c) From a room containing 13 people, choose a team of 5 people and designate one as a team captain. In how many ways can this be done?
Solution: C(13,5)*5 = 13*C(12,4) (d) From a pool of 17 girls and 10 boys, how many ways can you create a team of 8 girls and 2 boys?
Solution: C(17,8) * C(10,2) (e) From a room filled with 17 people, how many ways can you create a team consisting of 3 or 4 people?
Solution: C(17,3) + C(17,4) 3.
Determine the number of different arrangements of AABBCCDDEE such that each of the following conditions holds. Each of (a), (b), and (c) is a separate and independent problem. (a) The two As appear next to each other. (3 points)
Solution: 9!/(2!2!2!2!) (b) The two As are separated. (3 points)
Solution: 8!/(2!2!2!2!) * C(9,2) (c) The four vowels (A, A, E, E) are all separated. (4 points)
Solution: 6!/(2!2!2!) * P(7,4)/(2!2!) 4.
Ten dogs come upon eight biscuits, and dogs do not share biscuits! (a) In how many different ways can the biscuits be consumed by the dogs, assuming the dogs are distinguishable but the biscuits are not? (5 points)
Solution: C(17,9) Let d(i) represent the number of biscuits eaten by the ith dog, where 0 ² i ² 10. Then we seek the number on non-negative interger solutions to d(1)+d(2)+…+d(10) = 8. (b) In how many different ways can the biscuits be consumed by the dogs, assuming both the dogs and the biscuits are distinguishable? (5 points)
Solution: 10^8 For each of the 8 biscuits, there are 10 possible
dogs that could have eaten the biscuit. 5.
(a) Stacey and Petra are among 10 different women standing in line to enter a theatre. There are exactly 2 people between Stacey and Petra. In how many ways can such a line-up occur? (5 points)
Solution: 2 * 7 * 8! Stacey (S) and Petra (P), together with the two people between them, make up a fixed unit. With the four people fixed through this requirement, there are six others to account for, or a total of 7 "spots" in the line-up. First, choose one of the seven spots for the 4-person unit. This can be done in 7 ways. Now, using the six other spots and the two between S and P, permute the 8 other people into these 8 spots. This can be done in 8! ways. Finally, recognize that P and S could change places, thus there are two ways for S and P to arrange each other within the four-person unit. (b) Generalize your solution to the problem above for Stacey and Petra being among n different women standing in line to enter a theatre, with exactly k people between Stacey and Petra. (5 points)
Solution: 2*(n-k-1)(n-2)! Stacey (S) and Petra (P), together with the k people between them, make up a fixed unit. With the k+2 people fixed through this requirement, there are n(k+2) others to account for, or a total of n-(k+2)+1 = n-k-1 "spots" in the line-up. First, choose one of the n-k-1 spots for the (k+2)person unit. This can be done in n-k-1 ways. Now, using the n-(k+2) other spots and the k between S and P, permute the n-2 other people into these n-2 spots. This can be done in (n-2)! ways. Finally, recognize that P and S could change places, thus there are two ways for S and P to arrange each other within the (k+2)-person unit. 6.
Two soccer teams play until one team scores 10 points. The judges write down on a score sheet a record of how the score
changes. For example, a score sheet might look like this: 1/0, 1/1, 1/2, 1/3, 1/4, 2/4, 2/5, 2/6, 3/6, 4/6, 5/6, 6/6, 7/6, 8/6, 9/6, 9/7, 10/7 How many different score sheets can be obtained?
Solution: 2[C(9,9)+C(10,9)+C(11,9)+…+C(18,9)] Think about this as a modification or application of a grid problem. A "path" passes through integer ordered pairs (x,y), starting at the origin (0,0) and ending at either (a) an ordered pair on the line y = 10, with x an integer ranging from 0 to 9, or (b) an ordered pair on the line x = 10, with y an integer ranging from 0 to 9. We just need to count the paths, using the grid-problem strategy, required to get to each of these ending points. The only way to get to (x,10) or (10,y), for x <= 9 and y <=9, is to get there from (x,9) or from (9,y), respectively. That is, once the value "10" is reached, there is no more moving along the horizontal line y=10 or the vertical line x=10. As soon as a team scores 10, the match is over. The picture below may help illustrate this.
1.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e). a. Replace a, b, and c in C(a,5) = C(5,b) + C(5,c) to illustrate Pascal's Formula.
Solution: a=6, b=5, and c=4, or a=6, b=4, and c=5. b. A relationship s(n) is described recursively as s(n) = 3*s(n-1) - 2n, with s(1) = 4. State numerical values for s(2), s(3), and s(4).
Solution: s(2)=8, s(3)=18, and s(4)=46 c. A group of 24 employees will be shuttled across town in three vans, including a blue van, a red van, and a white van. How many different ways can the employees shuttle across town if 12 people are in the blue van, 7 are in the red van, and 5 are in the white van?
Solution: (24!)/(12!7!5!) d. A shelf is to contain 12 books, 8 indistinguishable paperback books and 4 indistinguishable hardback books. If the paperback books must be shelved in pairs (that is, exactly and only two paperback books must be adjacent to each other), in how many ways can the 12 books be arranged on a shelf?
Solution: C(5,4) e. State the explicit formula for determining the number of derangements of n objects.
Solution: D(n)=n!( (1/0!) - (1/1!) + (1/2!) - (1/3!) + ... + (-1)^n(1/n!) ) 2.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e). a. Determine the number of collected terms in the expansion of (2a-7c)^n.
Solution: n+1 b. Determine the value of the coefficient R in the collected
term Ra^5bc^12 resulting from the expansion of (a+b+c)^18.
Solution: (18!)/(5!1!12!) c. Determine the number of uncollected terms in the expansion of (s+t+a+r)^101.
Solution: 4^101 d. Determine the number of collected terms in the expansion of (b+r+a+i+n+s)^16.
Solution: C(21,5) e. Consider Pascal's Triangle, where 1 is the 0th row, 1 1 is the 1st row, and 1 2 1 is the 2nd row. i. Show the elements in row 5 of Pascal's Triangle.
Solution: 1 5 10 10 5 1 ii. State the difference between the sum of the elements that appear in row 15 of Pascal's Triangle and the sum of the elements that appear in row 16 of Pascal's Triangle.
Solution: 2^15 3.
After updating my advising records this week, I have found that 54 of my advisees each have accumulated from 54 credit-hours to 100 credit-hours. Explain why at least two of these 54 advisees must have the same number of accumulated credit-hours.
Solution: Here we assume that credit-hours accumulate in whole number increments. The range of values, from 54 to 100 credit-hours, spans 47 integer values: 54, 55, 56, ..., 98, 99, 100. The first 47 advisees may all have different numbers of accumulated credit-hours, but the 48th advisee is assured to match at least one of the first 47. This is justified by the pigeonhole principle. 4.
Gary drives a city taxicab during the summer. A client of his, Patti, always rides home in Gary's taxicab. Her apartment is 14 blocks north and 6 blocks east of the office building where she works. The streets between the office building and Patti's apartment are laid out in a rectangular grid, and all streets are accessible to Gary's taxi. How many different paths are available for Gary to make the 20-block trip from Patti's
workplace to her apartment?
Solution: This is a basic grid problem: C(20,14)=C(20,6). 5.
Thirty-two (32) distinct fair dice are rolled. How many ways are there for nineteen 5s to appear?
Solution: Because the dice are distinct, we must first select 19 of them as those that show a 5. This can be done in C(32,19) ways. Next we look at all ways that the remaining 13 dice can be rolled. There are 5 choices for the first die, 5 for the next, and so on, resulting in 5^13 ways for the remaining 13 dice to appear. Because each of the C(32,19) is a unique set of 19, each is paired with the 5^13 ways for the remaining dice to appear. This results in C(32,19)*5^13 ways for the nineteen 5s to appear. 6.
After administering a new medicine, a collection of 314 lab rats was tested for four diseases. One-hundred fifty-three (153) of the rats tested positive for asefachia, 179 tested positive for bunkeritis, 148 tested positive for cluenegligencia, and 155 tested positive for dipchillase. Among the same 314 rats, 85 tested positive for both asefachia and bunkeritis, 71 tested positive for both asefachia and cluenegligencia, 75 tested positive for both asefachia and dipchillase, 85 tested positive for both bunkeritis and cluenegligencia, 90 tested positive for both bunkeritis and dipchillase, and 77 tested positive for both cluenegligencia and dipchillase. We also know that 38 tested positive for all three of asefachia, bunkeritis, and cluenegligencia, 41 tested positive for all three of asefachia, bunkeritis, and dipchillase, 34 tested positive for all three of asefachia, cluenegligencia, and dipchillase and 47 tested positive for all three of, bunkeritis, cluenegligencia, and dipchillase. Finally, we know that 17 of the 314 lab rats tested positive for all four of the diseases. How many of the 314 lab rats tested negative for all four of the diseases?
Solution: Use the Inclusion/Exclusion principle: 314 - (153+179+148+155) + (85+71+75+85+90+77) (38+41+34+47) + 17=19 Nineteen (19) of the 314 rats tested negative for all four diseases. 7.
President John Kennedy was well known for his heroic efforts in World War II. During his career in politics, including service in Congress and as President, he developed a reputation for using war stories in his speeches. In fact, those who knew him best claim he told exactly 4 war stories per speech. They also claim that, in the 5004 speeches in which he told exactly 4 war stories, he never once repeated the same 4 stories in the exact same order. What is the minimum number of war stories Kennedy must have had at his disposal in order for those claims to be true?
Solution: We need to determine n so that P(n,4) is greater than or equal to 5004. By trial and error, n=10. Kennedy needed 10 war stories. 8.
Illinois State University surveyed parents/guardians and new students who took part in the university's "preview" activities. If the "preview" office kept track of whether each response was from a male parent/guardian, a female parent/guardian, a male new student, or a female new student, and the "preview" office received a total of 208 responses, how many different sets of 208 responses were possible, with respect to the gender-parent/student make-up of the respondents?
Solution: We seek the number of non-negative interger solutions to the equation MP+FP+MS+FS=208, where each two-letter variable represents one of the four categories of respondents. The number of such solutions is C(208+4-1,4-1)=C(211,3). 9.
A strange bequest by a recently deceased mathematician involved the two nephews of the mathematician. The two nephews were to divide the mathematician's sports card collection, with the only requirement being that each nephew get at least one card. If there were k different cards in the collection, with no card appearing more than once, how many different divisions were possible between the two nephews?
Solution: This problem can be viewed as determining the number of subsets from a kelement set and adjusting for the requirement that each nephew get at least one card. There are 2^k different subsets, including the empty set. We must eliminate both the empty set and the set itself as potential subsets, for use of either of these violates the restriction that each nephew get at least one card. Therefore, there are 2^k - 2 ways to divide the cards. 10.
An unlimited supply of dominoes is available for building a rectangle that measures exactly 2 units high and n units long. Each domino is a 2-by-1 rectangle. Write a recursive description for R(n), the number of different ways to build a 2-by-n rectangle with dominoes. Be sure to state any initial conditions of the relationship and to explain the basis for your recursive formula.
Solution: Let us look at some of the initial values for R(n). R(1) represents the number of 2-by-1 rectangles we can create with 2-by-1 dominoes. There is just one way to do this, so R(1)=1. R(2) represents the number of 2-by-2 rectangles we can create with 2-by-1 dominos. There are two way to do this, with a pair of dominoes both horizontal or both vertical, so R(2)=2. Now let's see how we can build R(3) from previous cases. From the 2-by-2 rectangles (i.e., those two in the R(2) case), we can append one vertical 2-by-1 domino. From the 2-by-1 rectangles (i.e., the one in the R(1) case), we can append two horizontal 2-by1 dominoes. It seems, then, that R(3) is just R(2)+R(1), or, equivalently, R(3)=3. This pattern will continue, for to get R(n) we need to just look at those in the R(n-1) case and append one vertical domino, or look at those in the R(n-2) case and appeand two horizontal dominoes. Therefore, R(n)=R(n-1)+R(n-2), with R(1)=1 and
R(2)=2. BONUS!
Each week two World Professional Tennis Organizations determine who are the #1 players in the world for both womens' and mens' professional tennis. I. A rare disease is known to affect 0.1% of the population. A test for the disease is 95% accurate. You test positive for the disease. What is the probability you actually have the disease? II. Each week two World Professional Tennis Organizations determine who are the #1 players in the world for both womens' and mens' professional tennis. Steffi Graf was #1 in womens' tennis without interruption over the period Monday, August 17, 1987, until Sunday, March 10, 1991. This is the greatest number of consecutive weeks that any woman or man has been #1 in the professional rankings. How many consecutive weeks were there in Steffi Graf's record?
1.
Pizza Hut is offering customers opportunity to create for themselves a unique music CD when they log on to the Pizza Hut website and use an appropriate code number. The Pizza Hut ad claims customers can make a 6-song CD by selecting from 200 different song titles. (a) How many different sets of 6 songs could be selected by a customer? (2 points) We seek the number of 6-song subsets from a set of 200 songs, where order or arrangement is not important . The solution is the number of combinations of 6 items selected from 200, or C(200,6)=(200!)/(194!*6!)=82,408,626,300. . (b) How many unique 6-song arrangements could a customer create? (2 points)
We seek the number of 6-song arrangements from a set of 200 songs, where order is important. The solution is the number of permutations of 6 items selected from 200, or P(200,6)=(200!)/(194!)=59,334,210,936,000.
Suppose that the 200 songs are listed in the following categories: Pop (44), Rock & Roll (52), Rythem and Blues (18), Jazz (21), Classical (36), and Show Tunes (29), with each song listed in one and only one
category. The numbers in parentheses above indicate how many songs are listed in each category. (c) If a customer selects one song from each category, how many sets of 6 songs could be selected by the customer? (3 points) We use the multiplication principle because we want one from each category to create a 6-song set. The wording of the problem ("sets of 6 songs could be selected") implies that order is not important. The solution is 44*52*18*21*36*29=902,918,016.
(d) Pat refuses to consider any Rock & Roll titles for her CD. How many 6-song CD arrangements does Pat have to choose from? (3 points) We seek the number of 6-song arrangements from a set of only 148 songs (200 total songs less 52 R & R songs). The wording of the problem ("6-song CD arrangements") implies that order is important. The solution is the number of permutations of 6 items selected from 148, or P(148,6)=(148!)/(142!)=9,484,150,515,840. 2.
Three friends each have a red, a white, a yellow, a blue, and a green Tshirt. (a) If each chooses a shirt to wear, how many unique 3-shirt sets could they be seen wearing? (5 points) We use the multiplication principle because we want one shirt on each person in order to create a 3-shirt set. There are 5 ways for each person to choose a shirt, and the choice one person makes does not alter the choice of anyone else. The solution is 5*5*5=125 unique 3shirt sets. . (b) If each chooses a shirt to wear, how many ways are there for each of them to all choose different colors? (5 points) We again use the multiplication principle because we want one shirt on each person in order to create a 3-shirt set. This time, there are 5 ways for the first person to choose a shirt, 4 for the next, and 3 for the third person. The number of choice diminishes because the second and third persons cannot choose a color that has already been chosen. The solution is 5*4*3=60 unique 3-shirt sets with each shirt a different color.
3.
Consider the letters in the word QUADRILLIONTHS. . (a) How many unique arrangements are there for the letters in this
word? (2points) There are 14! ways to arrange the 14 letters in the word, but two of the letters appear twice each. We account for the duplication of these letters by dividing by 2!*2!. The solution is (14!)/(2!2!1!1!1!1!1!1!1!1!1!1!)=(14!)/(2!2!)=21,794,572,800.
(b) How many arrangements exist if the four-letter sequence QUAD must be kept together in the order it now appears? (2 points) Consider the chunk QUAD as one unit. We now have 11 units to permute. There are 11! ways to arrange the 11 units in the word, but two of the letters appear twice each. We account for the duplication of these letters by dividing by 2!*2!. The solution is (11!)/(2!2!1!1!1!1!1!1!1!1!1!1!)=(11!)/(2!2!)=9,979,200.
(c) Considering only unique letters (no repetition), how many 10-letter subsets could be created from these letters? (3 points) There are 12 unique letters, so we seek the number of 10-letter subsets that can be created from a set of 12 elements. Order is not important. This is just a combination of 10 from 12, which is C(12,10)=(12!)/(2!*10!)=66.
(d) How many arrangements exist if each must begin and end with a vowel? (3 points) Within the 14 places for letters, place the first and last to assure they are vowels. There are 5 vowels to choose from for the first and 4 for the last. This is P(5,2)=5*4. Now place the remaining 12 letters. This can be done in P(12,12)=12! ways. Because each of the P(5,2) possible first-last vowel arrangements can be matched with the P(12,12) ways to place the remaining letters, we multiple the two results. Finally, we account for the duplication of two of the letters, requiring us to divide out those equivalent groups. We divide by (2!*2!) as we did in (a) and (b) above. This results in
[P(5,2)*P(12,12)]/(2!*2!)=(5*4*12!)/(2!*2!)=2,395,008,000 ways to rearrange the letters with a vowel in the first place and a vowel in the last place. 4.
At the Westminister Kennel Club dog show, held this month in Madison Square Garden in New York City and broadcast on the USA channel, every entrant must be assigned a unique registration number. One suggested strategy for assigning registration numbers requires a code that has three parts to it.
•
•
•
The first digit ranges from 1 through 7, corresponding to one of the seven dog groups the entrant is classified in. These are the herding group, the sporting group, the working group, the toy group, the terrier group, the hound group, and the non-sporting group. The next two digits correspond to the breed of the dog within the particular group. An English Cocker Spaniel, for instance, within the sporting group, is breed 14. No group has more than 27 breeds. The final numbers in the registration code provide more information about the particular dog. The first of the final digits is a 0 or a 1, based on the sex of the dog. The second and third digits in this last set represent the age of the dog in years, starting with 00 if less than 9 months old and ranging up to 12, . the oldest dog in this year's show being 12 years old. The last two digits in this final portion of the registration code represent the state of residence of the dog. Values 01 through 50 represent the 50 United States. If a dog is from outside the U.S., this number appears as 00.
(a) Without considering any other circumstances or restrictions, how many unique registration codes are possible under this scheme? (8 points)
Consider the registration number and the number of values that could occur in the various positions. We then use the multiplication principle to determine the total number of possibilities because we want values in all the positions.
This results in 7*27*2*13*51=250,614 possible registration codes.
(b) What problems in dog identification could occur under this strategy? (2 points) One problem is that two unique dogs in the show may be assigned the same registration number. Two or more dogs may have exactly the same characteristics, as described by the different criteria in the suggested registration-number assignment scheme described above. There would be no way, by this scheme, to distinguish between the two dogs. 5.
In a certain leap year, the 13th of the month was a Friday three different times. What day of the week was 29 February that year? (10 points)
One way to approach this problem is to label the days of the leap years using positive integers from 1 through 366. The table below shows the labels for the 13th day of each month for leap years. In parentheses after a 13th's label is the remainder when the label is divided by 7. Call this an index number for the 13th of each month. Month
Leap-Year Labels and Index Numbers for 13th of Each Month
January
13 (6)
February
44 (2)
March
73 (3)
April
104 (6)
May
134 (1)
June
165 (4)
July
195 (6)
August
226 (2)
September
257 (5)
October
287 (0)
November
318 (3)
December
348 (5)
The only index number that occurs three times is 6. This means the index number 6 represents Fridays. •
We determine which day of the week the 29th of February
.
•
•
lands in one of several ways. One way is to move forward from the 13th of February. This has index number 2. That means the 20th of February has index number 2 as does the 27th of February. That means the 29th of February has index number 4. If index number 6 is Friday, index number 4 is two days earlier, or Wednesday. Another way is to move back from the 13th of March. The 13th of March has index number 3, so the 6th of March and the 28th of February will have index number 3. The 29th of February will have index number 4, a Wednesday. A third way is to use the leap-year label for the 29th of February. That label is 60 (31 days in January and 29 in February). We divide 60 by 7 and the remainder, 4, is the index number for the 29th of February. As we have already determined, index number 4 represents a Wednesday.
When a leap year has three Friday the 13ths, the 29th of February will occur on a Wednesday. 6.
Roberta claimed that the following equation was always true for positive integers n > 1.
(a) Is Roberta correct? (2 points) Yes!
(b) If Roberta is correct, justify her result for the general case. If Roberta is not correct, provide an example to show she is not correct (called a counter example). (8 points) To justify Roberta's claim in the general case, we need to show that the left side and right side of her equation are equivalent. Here's one way to do that.
In the first line above, we have expanded n! into its factors. In the second line, we rewrite (n-2)(n-3)(n-4)...(2)(1) in its equivalent form,
(n-2)!. In the third line, we expand n(n-1) to get n^2-n. The third line is exactly the right side of Roberta's equation. Therefore, we have shown how to correctly and logically move from the left side of Roberta's equation to the right side of Roberta's equation. That shows us that the left side and right side are equivalent, which in turn justifies that the equation holds in general.
1.
Respond to each of these questions. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) on this page. a) Express P(16,5) using factorial notation.
b) How many distinct arrangements exist for the letters in the word reverberator ?
c) In the expansion of
, state:
i) the number of uncollected terms: 4^12 ii) the coefficient J in the collected term
:
d) Determine the number of collected terms in the expansion of
.
6 e) Replace w, x, y, and z in C(12,6) = C(w,x) + C(y,z) to illustrate Pascal's Formula, a fundamental relationship that exists in Pascal's Triangle. C(12,6) = C(11,6) + C(11,5) 2.
Thum lives in Grid City, where the streets are laid our in a grid, running east/west and north/south. Thum's house is in the northwest corner of the city and his girlfriend, Bolina, lives in the southeast corner of the city. Thum's house is 8 blocks north and 12 blocks west of Bolina's. The image below shows the entire city.
a) How many 20-block paths are there from Thum's to Bolina's, assuming all streets exist and are open to traffic? Traveling from Thum's to Bolina's in a 20-block path requires traversal of 8 southbound streets and 12 eastbound streets. Thus, we need to look at all permutations of the "word" SSSSSSSSEEEEEEEEEEEE to determine the number of paths. The number of unique arrangements of this 20-character word is C(20,8)=C(20,12)=20!/(12!8!). b) Grid City eventually will build a walking mall in the shaded location shown below, thereby eliminating a one block length of street. Under these conditions, how many 20 block paths are there from Thum's to Bolina's?
From the C(20,8) paths determined in (a), we must subtract all paths that contain the restricted street, shown above as going from x to y. Using the same reasoning as in (a), there are C(9,4) paths from T to x, 1 path from x to y, and C(10,4) paths from y to B. Because each of these legs (T to x, x to y, and y to B) are independent from each other and each must be included in a 20-block path from T to B through the restricted street, we multiply the three values to determine the total number of 20-block paths from T to B through the restricted street. This is C(9,4)*1*C(10,4)=C(9,4)*C(10,4). We subtract this from C(20,8) to determine the number of paths we can use to go from T to B and avoid the restricted street. The resulting number of paths is C(20,8)-C(9,4)*C(10,4). 3.
Referring to the letters in the word ACMAESTHESIA, solve each of the following problems. Each problem is independent and separate from the others. a) How many unique arrangements are there for the letters in this word? There are 12! permutations of the letters in the word, but we must account for duplication and therefore divide by 3!2!2! because there are 3 identical As, 2 identical Es, and 2 identical Ss. There are 12!/(3!2!2!) unique arrangements for the letters in this word. b) How many unique arrangements exist if two or more vowels cannot be adjacent to one another? Begin by placing the 6 consonants in the word. Accounting for two Ss, this can be done in 6!/2! ways. This leaves us 7 places among the 6 consonants in which to place vowels so no vowels will be adjacent. We first choose 6 of the 7 places for the vowels, done in C(7,6) ways. We now
permute the vowels within the places chosen. this can be done in 6!/(3!2!) ways, accounting for duplicate As and Es. We multiply the results to determine the total number of ways of arranging the letters in the word so no vowels are adjacent. This is (6!/2!)*C(7,6)*[6!/(3!2!)]. c) If the only distinction we can make is between consonants and vowels, how many arrangements can be made? With no distinction among the vowels or among the consonants, we have 12 total objects of just two types, 6 of type "consonant" and 6 of type "vowel." We permute the 12 objects and account for the duplicates among the types. This results in 12!/(6!6!) ways to make this sort of arrangement. d) If the only distinction we can make is between consonants and vowels, and two or more consonants cannot be adjacent to one another, how many arrangements can be made? Again with no distinction among the vowels or among the consonants, we have 12 total objects of just two types, 6 of type "consonant" and 6 of type "vowel." We first place the vowels, and because they are indistinguishable, there is 1 way to do that. We create 7 spaces among the vowels, and we choose 6 of these for the consonants. This is just C(7,6), our desired result. There are 7 arrangements that can be made under these conditions. 4.
Louise invests her money in $200 lots. She has $3000 to invest and her daughter Gina has suggested five different mutual funds for Louise's investments. a) How many different ways can Louise invest her money if she insists on putting at least $200 in each of the five funds her daughter recommended and uses only these five funds? Note that Louise has 15 lots of $200 each to invest. Thus we are dealing with the placement of 15 objects that are identical. If Louise drops 1 lot into each of the five funds, she has 10 lots remaining to distribute. This distribution of 10 lots can be done in any way among the five funds. This is analogous to solving the equation A+B+C+D+E=10 where A,B,C,D, and E must be nonnegative integers. This
placement can be done in C(10+5-1,5-1)=C(14,4) ways. In terms of objects and dividers, we have 10 objects and 4 dividers to permute. This can be done in 14!/(10!4!) ways. b) If Louise restricts her investments to these five funds but may choose to not invest any money in one or more of the funds, how many different ways can Louise invest her money? Again we have 15 lots of $200 each to invest. This distribution of 15 lots can be done in any way among the five funds. This is analogous to solving the equation A+B+C+D+E=15 where A,B,C,D, and E must be nonnegative integers. This placement can be done in C(15+5-1,5-1)=C(19,4) ways. In terms of objects and dividers, we have 15 objects and 4 dividers to permute. This can be done in 19!/(15!4!) ways. 5.
Consider the binomial
.
a) Determine the number of uncollected terms in the expansion of this binomial. For the expansion of a binomial, this is just 2^t. b) Show and describe a typical collected term in the expansion of this binomial.
A typical term in the expansion will be where x+y=t and C=t!/(x!y!).
,
c) If k = x, m = 2y, and t=8, determine N in the collected term Nx^3y^5. We are expanding the binomial (x+2y)^8. for the collected term in question, we will have C(8,3)*(x)^3*(2y)^5, which
simplifies to
. Therefore,
. 6.
A bag contains a virtually unlimited supply of red marbles, blue marbles, white marbles, and yellow marbles. Marbles of any one color are indistinguishable from each other. a) Marbles are drawn from the bag without looking until a set of 6 marbles is created. How many different 6-marble sets could be created? We have an essentially unlimited supply of four different objects, so as we draw from the bag we could generate a set of 6 marbles with various numbers of each color marble, as many as 6 of one color and as few as 0 of one or more colors. This is analogous to solving the equation R+B+W+Y=6 where R,B,W, and Y must be nonnegative integers. This distribution can occur in C(6+4-1,4-1)=C(9,3) ways. In terms of objects and dividers, we have 6 objects and 3 dividers to permute. This can be done in 9!/(6!3!) ways. b) Marbles are drawn from the bag without looking until a set of 24 marbles is created. Of all the 24-marble sets we could create, how many have at least two marbles of each color? Again we have an essentially unlimited supply of four different objects, so as we draw from the bag we generate a set of 24 marbles with various numbers of each color marble. We are restricted to having at least two of each color marble. If we preselect 2 marbles of each of the four colors, we have 16 marbles remaining to select. This selection of 16 marbles can be done in any way among the four colors. This is analogous to solving the equation R+B+W+Y=16 where R,B,W, and Y must be nonnegative integers. This distribution can occur in C(16+4-1,4-1)=C(19,3) ways. In terms of objects and dividers, we have 16 objects and 3 dividers to permute. This can be done in 19!/(16!3!) ways.
c) Marbles are drawn from the bag without looking until a set of 30 marbles is created. Of all the 30-marble sets we could create, how many have marbles of exactly three colors in them? Again we have an essentially unlimited supply of four different objects and we draw from the bag to generate a set of 30 marbles with various numbers of each color marble. We are restricted to 30-marble sets that contain exactly three colors. This is equivalent to saying in the 30marble set exactly one color is NOT represented. If we preselect red as the marble color not selected, we have 30 marbles to select from among three colors. This selection of 30 marbles can be done in any way among the three colors, with the restriction that each of the remaining three colors is represented by at least one marble. This is analogous to solving the equation B+W+Y=30 where B,W, and Y must be positve integers. This distribution can occur in C(30-1,3-1)=C(29,2) ways. In terms of objects and dividers, we have 30 objects and 2 dividers to work with. Because all remaining colors among the three must be represented, the 2 dividers must be placed somewhere among the 29 spaces between the 30 objects. This can be done in C(29,2) ways. The same logic and calculations will result for each of the other three cases, that is, when blue is the preselected absent color, when white is the preselected absent color, and when yellow is the preselected absent color. Therefore, of all the 30-marble sets we could create, there are 4*C(29,2) with exactly three colors in them.
1.
Respond to each of these questions. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (d) of problem 1.
a) Determine the number of derangements of the elements in the set {1,2,3,4,5,6}, where the natural position for 1 is the first position, the natural position for 2 is the second position,
and so on. b) Answer the following questions for the recurrance relationship described by: a(n)=5a(n-1)&endash;2a(n-3) for n a positive integer, where a(1) = &endash;5, a(2) = &endash;6, and a(3) = &endash;1. (i) Determine a(6). (ii) Determine the smallest value of n such that a(n+1) &endash; a(n) is greater than 5000. c) In the expansion of
, state:
i) the number of uncollected terms ii) the coefficient K in the collected term
.
d) Determine the number of collected terms in the expansion of 2.
.
Georgania Higgenbottom served a variety of cool drinks at her spring garden party. Each of the 120 guests selected exactly one beverage, chosen from the following list: ginsing go-getter, spinach surprise, cabbage combo, tomato torment, and cucumber delight. Georgania's head waiter, Christof, kept a tally sheet of drink selections for the 120 guests. a) Suppose Christof noticed that each of the beverages had been selected by at least one guest. How many different tally sheets could Christof have generated for the party's drink selection under these circumstances? b) Suppose instead that when Georgania screamed to Christof, "What are our minimums on drinks?" he replied, "I see that at least 10 people chose the ginsing go-getter, 20 or more went for the spinach surprise, more than 6 people tried the cabbage combo, no one selected tomato torment, and at least 15 drank cucumber delight." Under these conditions, how many different tally sheets could Christof have generated for the party's drink selection?
c) Try to picture Amiele, the beverage host, walking around
the table asking guests for their beverage choice. How many guests would Amiele have to ask before he would be assured that a beverage choice repeated itself? What assumptions underlie your response? 3.
In the Ancient yet little-known game of gnikoj, scoring occurs several ways. A player can earn 1 point for a kaboom, 1 point for a laboom, and 1 point for a maboom. A player earns 2 points for executing a bifter or for executing a cifter. In traditional gnikoj, there is no way to earn 3 points on a single move, and the only way to earn 4 points is by the rarely seen move called a gemserp . Points are added and the player with the highest point total wins. a) Show at least three different scoring sequences a gnikoj player could execute and earn 11 points. Note that the order in which points are earned is significant. A kaboom followed by a cifter is a different scoring sequence than a cifter followed by a kaboom. b) In gnikoj, play continues until an essapmi is reached. Individual player totals could exceed 100 points. Create a recursion relationship T(n), including any required initial conditions, to describe the number of different sequences of gnikoj moves that result in a total of n points. Show justification for your result.
4.
What is wrong with the following problem situation? A survey of 144 new teachers in a metropolitan school district determined that 16 new teachers had been to Europe, 13 had been to Asia, and 17 had been to Africa. Of those who had been to Europe, 10 had been to Africa. Of those who had been to Asia, 9 had been to Africa. Of all the new teachers surveyed, 5 had been to Africa, Asia, and Europe, and 111 new teachers had been to none of those continents.
Describe specifically and clearly what is wrong and how you determined that. 5.
a) Using an unlimited supply of letters from the set {A,B,C,D}, how many 7-letter sets are possible to create with at least one C in the set? b) Using an unlimited supply of letters from the set {A,B,C}, how many 5&endash;letter words (meaningless or not) are possible to create with exactly one A in the word?
6.
In a local chess tournament, two players compete until one player wins 4 matches. With no regard for the ability of the
players, what fraction of all possible two-player competitions will require exactly 7 matches?
1.
a. Show how to simplify the following expression to generate a positive integer: C(5,2).
b. Determine the number of ways to arrange 10 distinct dogs in a straight line. There are 10! arrangements. c. There are 8 coffee choices and 12 tea choices on the menu at Farms and Babble Bookstore. These are the only beverage choices. (i) If a customer orders either tea or coffee, how many selections does the customer have to choose from? In this case, using the Addition Principle, there are 8 + 12 = 20 selections. (ii) If a customer orders one tea choice and one coffee choice, how many choices are possible? Disregard whether tea or coffee is ordered or served first. Using the Multiplication Principle, there are 8 x 12 = 96 choices. d. Determine the number of non-negative integer solutions to the equation A + B + C + D + E + F = 40. C(40+6-1,6-1) = C(45,5). e. Consider Pascal's Triangle, where 1 is the 0th row, 1 1 is the 1st row, and 1 2 1 is the 2nd row. (i) Show the elements in row 6 of Pascal's Triangle. Row 6: 1 6 15 20 15 6 1
(ii) State the sum of the elements that appear in row 20 of Pascal's Triangle. The sum is 2^20. 2.
a. Determine the number of collected terms in the expansion of
. There are w+1 collected terms.
b. Determine the value of the coefficient K in the collected term resulting from the expansion of
. K = 9!/(2!7!) c. Determine the number of uncollected terms in the expansion of
. There are 4^15 uncollected terms.
d. Determine the number of collected terms in the expansion of
. There are C(22+5-1,5-1) = C(26,4) collected terms.
3.
The following conjecture is to be proven true by induction or shown to be false using a counterexample:
a. State and carry out the first step in the induction process.
Step 1: Show that the conjecture is true when n=1. When n=1, the LS is (2-1)(2x1)=2. When n=1, the RS is (1x3x2)/3=2. When n=1, LS=RS. b. State and carry out the second step in the induction process. Step 2: Assume that the conjecture holds when n=k. When n=k, the conjecture is
c. State but do not carry out the third step in the induction process. Step 3: Use the assumption in Step 2 to show that the conjecture holds when n=k+1. 4.
A passenger jet may fly three routes from New York to Chicago and four routes from Chicago to Los Angeles. For a round trip from New York to Los Angeles and back, determine the number of ways a passenger can travel without repeating the same route on any leg of the round trip. Starting in NY, there are 3 ways to get to Chicago and 4 ways to get from Chicago to LA. Therefore there are 3x4=12 different ways to get from NY to LA via Chicago. On the return trip, the passenger has only 3 ways to go from LA to Chicago because a previously used route cannot be repeated. Likewise there are 2 ways to go from Chicago back to NY. Therefore, the return trip has 2x3=6 ways to return to NY from LA via Chicago. By the multiplication principle, there are 12x6=72 different round trips in all that reuse none of the legs of the journey.
5.
Five unique dice are thrown simultaneously. Determine the portion of all possible throws that result in at least two 5s appearing. First determine the number of different throws that are possible with no restrictions: There are 6 ways for each die
to land, so there are 6x6x6x6x6=6^5 unique throws. Now determine how many throws show at least two 5s. do this by subtracting from 6^5 the number of ways to get exactly one 5 and exactly no 5s. To get no 5s, there are 5x5x5x5x5=5^5 ways for the dice to land, because each die can result in one of five outcomes (everything but 5). To get exactly one 5, there are 5 ways to choose which die will be the 5, and then once that choice is made, the remaining four dice each have 5 outcomes possible. Therefore, there are 5(5x5x5x5)=5^5 ways for exactly one 5 to appear. Therefore, there are 5^5 + 5^5 ways to get exactly zero or one 5 to appear, so there are 6^5-(2x5^5) ways for at least two 5s to appear. Of all possible throws, the portion that show at least two 5s when five unique dice are thrown is [6^5-(2x5^5)] / 6^5. 6.
A nurse walks from home at 10th and H to her clinic at 16th and M, always walking to higher numbers or to letters further along in the alphabet. On a certain day, police block off 13th street between K and L streets. What portion of all possible paths from the nurse's home to the clinic contain the blockedoff street? First determine the total number of ways to walk from 10th and H to 16th and M with no restrictions. Assume that the numbered streets run east and west and the lettered streets run north and south. Also assume that the numbered streets get higher as you move south and the lettered streets get further into the alphabet as you move east. The nurse must walk 6 blocks south and 5 blocks east. There are C(11,6) ways to do this. Now determine how many of these paths include the blocked-off street. To get to K and 13th, you must walk 3 blocks east and 3 blocks south. Therefore, there are C(6,3) paths from 10th and H to 13th and K. There is 1 way to get from 13th and K to 13th and L. From 13th and L, there are 3 southerly blocks and 1 easterly block to walk to get to 16th and M. There are C(4,3) walks to walk this stretch. All in all, there are C(6,3)x1xC(4,3) paths from 10th and H to 16th and
M that include the restricted street. We remove these from the total number of paths, yielding C(11,6) - [C(6,3)x1xC(4,3)] total paths that do not include the blocked-off street. 7.
A company named GAMES has an advertising display with the letters of its name, "GAMES." Colors are used for each letter, but the colors may be repeated. On one particular day, for example, the colors might be red, green, green, blue, red. The company wishes to use a different color scheme for each of the 365 days in the year 2001. Determine the minimum number of colors that are required for this task. One way to approach this problem is to think of the choices available to fill each of 5 available spaces. Here those 5 spaces are the letters, and the things we fill them with are colors. If there is 1 color available, we have 1 way to fill the first space, 1 way to fill the second, and so on. Repetition is allowed. Using the multiplication principle, there is 1x1x1x1x1 way to color the letters in the sign. If there are 2 colors available, we have 2 ways to fill the first space, 2 ways to fill the second, and so on. Repetition is allowed. Using the multiplication principle, there are 2x2x2x2x2=2^5 ways to color the letters in the sign. In general, there are n^5 different coloring schemes, where n is the number of available colors. We seek n such that n^5 >= 365. We have 1^5=1, 2^5=32, 3^5=243, and 4^5=1024. The minimum number of colors is 4.
8.
In 7,843 families, all of which have a TV set, a dishwasher, a microwave, and a car, there are six different types of TV sets, five different types of dishwashers, four different types of microwaves, and eight different types of cars. What is the least number of families that have the same type of TV, dishwasher, microwave, and car? By the multiplication principle, there are 6x5x4x8=960 different combinations of the 4 objects when we consider the
different number of types of each object. By the pigeonhole principle, we may have the first 960 families surveyed each identify a different combination of the 4 objects, but by the 961st family, we will repeat one of the 960 combinations. Dividing 7843 by 960, we get a quotient between 8 and 9. Thus we could have 8 full sets of the 960 pigeonholes, and some with at least 9 responses. Thus, there are at least 9 families that share the same type of objects. 9.
Cannon balls are stacked in a compact equilateral triangular pattern. When there are n layers in the stack, there are n balls per side of the triangle on the lowest layer, n-1 per side on the next layer, and so on, up to 1 ball on the top. Determine a recursion relationship B(n), including any initial conditions, for the total number of balls in a pile with n layers. Here is a two-dimensional visual image of several layers, moving down from the top.
Notes: •
•
•
Each layer represents atriangular number: 1,3,6,10, and so on. These numbers each represent the sum of a specific number of positive integers: 1=1, 3=1+2, 6=1+2+3, 10=1+2+3+4. Recall that the sum of the first n positive integers is [n(n+1)]/2.
Now consider some initial examples of the cannon-ball stacks. • •
A stack with one layer has 1 cannon ball, so B(1)=1. A stack with two layers has layers (a) and (b) shown
•
•
above, so it has 4 cannon balls. Another way to think of this is that it has the number B(1) plus the new layer, here figure (b). So B(2)=B(1)+3. A stack with three layers has layers (a) through (c) shown above, so it has 10 cannon balls. Another way to think of this is that it has the number B(2) plus the new layer, here figure (c). So B(3)=B(2)+6. A stack with four layers has layers (a) through (d) shown above, so it has 20 cannon balls. Another way to think of this is that it has the number B(3) plus the new layer, here figure (d). So B(4)=B(3)+10.
So what do these examples illustrate in general? Each new pile of cannon balls has the same number as the previous pile plus the number in the lew layer. That new layer will always have a number of cannon balls equal to the sum of the positive integers up to that layer number. That is, for a stack with n layers, the number of balls will be B(n-1) plus the sum of the first n positive integers. In compact notation, we have B(n)=B(n-1)+[n(n+1)]/2, with initial condition B(1)=1. 10.
Exactly 10 chocolate chips are to be distributed at random into 6 chocolate-chip cookies. What is the probability that some cookie has at least 3 chips in it? We need two values to compare as a ratio to determine the desired probability: The total number of ways to distribute 10 chocolate chips into 6 cookies and the number of those ways that result in at least three chips in one cookie. To determine the total number of ways to distribute the chips, we determine the number of non-negative solutions to the equation A+B+C+D+E+F=10, where each letter represents a cookie. The number of solutions is C(10+6-1,6-1)=C(15,5). To determine the number of those ways that result in at least three chips in one cookie, we solve the complementary problem and subtract that result from the total number of ways. The complementary problem is to determine the number of ways that the chips can be distributed so less than three chips
are in any cookie. Considering the equation above, we want A,B,C,D,E, and F to be values less than 3. One case in this solution is for a cookie to have no chips in it. If that is the case, the other five cookies will each have two chips. There are 6 cookies to choose from in determining which will have no chips, so there are 6 ways to distribute the chips so one cookie gets no chips. We can also distribute the chips so two cookies get one chip each and the rest get two each. This can be done in C(6,2) ways, for we need to determine the number of ways to select two of the six cookies into which we place one chip each. There are no other ways to restrict the chips to less than three chips, for there is no way to have two or more cookies with no chips (without some of the others having at least three chips), and if more than two cookies each have one chip, we are assured of at least one cookie getting three or more chips. So there are 6+C(6,2)=6+15=21 ways to restrict chip distribution to two chips or less. That means there must be C(15,5)-21 ways to have three or more chips in a cookie. This means the desired probability is [C(15,5)21]/C(15,5)=2982/3003 or approximately a 99.3% probability that a cookie will get three or more chips. BONUS!
A survey was conducted of 983 families to determine whether they possessed (1) a cell phone, (2) a microwave, (3) a satellite dish, or (4) a CD player. No family was completely without such items, and 481 families had at least two of these items. At least three items were possessed by 345 families and 264 families possessed all 4 items. a. Determine the total number of pieces of equipment held by all 983 families. b. Determine the number of families that held the number of pieces of equipment specified below. Assume that none of these families had more than one of any particular item. (i) exactly one piece of equipment (ii) exactly two different pieces of equipment (iii) exactly three different pieces of equipment
1.
A bus filled with a high school music group stopped at Blaise's Bistro. The director, a budding mathematician, noticed the menu board at the Bistro and quickly assured her assistant that at least one of the 8 sodas listed on the board would be ordered at least 4 times by the student musicians. Naturally, the director assumed that each student would order exactly one soda from the list. What is the minimum number of students that must be in this music group?
.
Suppose that none of the 8 sodas is ordered four times or more. It is possible that all 8 varieties were ordered three times each, requiring 24 students. A 25th student, however, would force the fourth order of one of the sodas, by the Pigeonhole Principle. Therefore, the minimum number of students in the group is 25. 2.
North American radio stations must adhere to specific guidelines when selecting the call letters for the station name. •
•
The name must contain either three or four letters of the alphabet. The name must begin with a W or a K.
How many different radio station names are possible under these restrictions? Consider two disjoint cases. Case I: 3 letters in the name In this case, there are 2 choices for the first letter and 26 choices for each of the second and third letters. Each choice is independent so by the multiplication principle, there are 2*26*26 possible 3-letter names.
.
Case II: 4 letters in the name In this case, there are 2 choices for the first letter and 26 choices for each of the second, third, and fourth letters. Each choice is independent so by the multiplication principle, there are 2*26*26*26 possible 4-letter names. Because the cases are disjoint, we add the results of the two cases to determine the total number of names that are possible. 2*26*26 + 2*26*26*26 = 1352 + 35,152 = 36,504 3.
Consider the letters in the word SIMULATE. . (a) How many rearrangements are there of these letters?
P(8,8) = 8! (b) How many rearrangements exist if the three-letter sequence SIM must be kept together as it appears? Consider the chunk SIM as one unit. We now have 6 units to permute. This is just P(6,6)=6! (c) How many rearrangements exist if each must begin and end with a vowel? Within the 8 places for letters, place the first and last to assure they are vowels. There are 4 vowels to choose from for the first and 3 for the last. This is P(4,2). Now place the remaining 6 letters. This can be done in P(6,6) ways. Because each of the possible first-last vowel arrangements can be matched with the P(6,6) ways to place the remaining letters, we multiple the two results. This results in P(4,2)*P(6,6) ways to rearrange the letters with a vowel in the first place and a vowel in the last place. (d) From the 8-letter set, how many 5-letter subsets exist? This is just a combination of 5 from 8: C(8,5). 4.
Using the set {A,B,C}, what fraction of all 5-letter words that can be created contain exactly one A? There are a total of 3*3*3*3*3 = 243 5-letter words that can be made. To determine the number of words with exactly one A, we note that there are 5 places for the one A. That is, an A can and must be placed in one of the five positions in the word. There remain 4 places and for each there are two choices. This results in 5*2*2*2*2 or 80 possible 5-letter words that contain exactly one A.
.
This results in 80/243 as the fraction of all 5-letter words, from the set {A,B,C}, that contain exactly one A. 5.
Al and Bobbie are in a group of 12 students. Three teams of 4 students are to be created from the group of 12. Among all possible three-team sets, how many ways exist for Al and Bobbie to be on different teams? There are 3*2 = 6 ways to place Al (A) and Bobbie (B) onto two different teams from the three teams being created. Although not necessary, you can think of the teams with names RED, BLUE, and GREEN, and count the possible ways A and B could be placed onto
.
two different teams. Now that each of A and B are in a designated team, choose from the remaining students to fill out the teams. From the 10 remaining students, choose 3 to fill out the team A is on. This can be done in C(10,3) ways. Now from the 7 remaining students, choose 3 to fill out the team B is on. This can be done in C(7,3) ways. Finally, place the remaining 4 students on the third team, the one that neither A nor B is on. This results in 6*C(10,3)*C(7,3) ways to meet the conditions of the problem. 6.
Complete ONE of the two problems listed below. I. Show an algebraic proof that C(a,c)*C(a-c,b-c) = C(a,b)*C(b,c) for a>=b>=c. The strategy to use here is to represent each combination expression using factorial notation and through simplification, show that the leftside product is equivalent to the right-side product.
The final expressions in [1] and [2] above are equivalent, thereby showing that C(a,c)*C(a-c,b-c) = C(a,b)*C(b,c). This completes the required algebraic proof. II. Determine the smallest natural number n that assures
We have shown in class that , so our goal is to determine the smallest natural number n so that . We first manipulate the inequality:
Using conventional methods for solving equations (complete the square or use the quadratic formula on [4], carry out a guess-and-test on [3], graph the expression on the LS of [4] and look for x-axis intercepts, use a CAS SOLVE routine, and so on), we determine that the required natural number n is 1414.
1.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) of question (1). a. Express C(10,4) in terms of P(10,6). C(10,4)=C(10,6), and C(10,6)=P(10,6)/6! so C(10,4)=P(10,6)/6!. b. How many distinct arrangements exist for the letters in the word ATTESTANT? 9!/(2!4!) c. In the expansion of (m + n + p)^10, determine the number of: (i) uncollected terms 3^10 (ii) collected terms C(12,2) d. Express C(n,k) + C(n,k + 1) as a single combination. C(n+1,k+1) e. Let S(k) represent the sum of the elements in row k of Pascal's Triangle. For example, S(0) = 1 and S(1) = 2. Express S(12) - S(10) in terms of S(10).
.
S(12)=2^12 and S(10)=2^10, so S(12)-S(10)=2^12-2^10=2^10(2^2-1)=3*(2^10)=3*S(10). 2.
Members of a mathematics department will take a one-day course on using the TI-89 calculator in collegiate mathematics instruction. All 36 instructors in the department must take the course and can take the course only once. The course will be offered on the Monday, Tuesday, Wednesday, and Thursday immediately after spring semester graduation. Faculty must sign-up in advance for the day they wish to complete the course.
.
In the following questions, we are not concerned about who among the faculty members take the course on a certain day, but about how many take the course each day. a. If there are no restrictions about how many take the course on any one day, how many ways exist for the department to sign up for the courses? C(36+4-1,4-1)=C(39,3) b. If at least one faculty member must sign-up for each of the days the course is offered, how many ways exist for the department to sign up for the courses? C(36-1,4-1)=C(35,3) c. Because of availability of course leaders, it is necessary to assure that certain enrollments are met for specific days the course is offered. Administrators determined that at least 5 faculty members must take the Monday course, at least 4 faculty members must take the course on Tuesday and at least 4 on Wednesday, and there must be more than 5 enrolled for the Thursday version. How many ways now exist for the department to sign up for the courses? With the stated restrictions, placement of 19 faculty is assured. This leaves 17 to distribute, with possibility that one or more days may have no more enrollees. C(17+4-1,41)=C(20,3). 3.
Members of a 7th-grade class recently identified their interests in collecting things. Among those enrolled in the class, 15 collect stamps, 14 collect coins, and 18 collect hobby cards. Exactly 8 of the students collect both stamps and coins, 7 of them collect both coins and hobby cards, and
.
9 of them collect stamps and hobby cards. In the class, there are 6 students who collect stamps, coins, and hobby cards. If at least one student enrolled in this 7th-grade class does not collect any of the items listed above, what is the smallest number of students enrolled in this class? Using a Venn diagram we can count those students accounted for according to the description of their collecting interests. With at least one student who does not collect any of the three items listed, there must be at least 30 students enrolled in the class. 4.
In the expansion of (a + b + e + d + e)^21, determine the number of different ways a coefficient of 21 appears among the collected terms. We want to determine the number of times K=21 when we examine the collected terms of the form Ka^Ab^Bc^Cd^De^E where A+B+C+D+E=21 over the nonnegative integers. We know that K=(21!)/(A!B!C!D!E!), the multinomial coefficient. For K to be 21, it must be the case that A!B!C!D!E!=20! [Why is that?] For that to be true, two conditions must be met: (i) We must have exactly one of A, B, C, D, or E equal to 20 and (ii) the remaining 4 values must sum to 1. There are 5 ways for the first condition to be met: A or B or C or D or E must be 20. There are 4 ways for the second condition to be met: one of the four remaining values from among A,B,C,D,E must be 1 and the other three must be 0. By the multiplication principle, there are 5*4=20 total ways for a coefficient of 21 to appear.
.
5.
How many ways can the letters in the word adequateness be arranged so that three consonants start the arrangement and three consonants end the arrangement? Here is a sequence of steps that develops the elements we need in order to solve the problem. •
•
•
• •
Choose and permute three of the six consonants, to be placed in front: P(6,3) Permute the remaining three consonants to be placed at the end: Multiply by P(3,3)=3! Correct for duplication caused by appearance of 2 Ss: Divide by 2! Permute the 6 vowels: Multiply by P(6,6)=6! Correct for duplication caused by appearance of multiple vowels: Divide by 2!3!
This results in this solution: [P(6,3)*3!*6!] / [2!2!3!] = [P(6,3)*6!] / [2!2!] 6.
My friend Michael Link posed this problem when we met as part of an NCTM committee. It seems that while at school one day, he met a locksmith who was trying to fix a door lock. The lock was a type you may have seen before, where the user presses buttons in a certain sequence to open the lock. This particular lock had four pads arranged vertically, each labeled with a different letter A, B, C, and D. A valid combination for the lock met the following restrictions: • •
• •
The combination must have exactly two inputs. An input means to press from one, two, or three letters simultaneously. No letter could be pressed more than once. Input order is important. For example, to press "A" and then the pair "B/D" represents a different combination than to press "B/D" and then "A." However, the single input "B/D" is the same as the single input "D/B" because the two letters are pressed simultaneously.
How many different combinations exist for the lock described here? From the restrictions on a valid combination that are provided with the question, there are exactly six different ways to create exactly two inputs, as shown in the table
.
below. 1 letter, 1 letter 1 letter, 2 letters 2 letters, 1 letter 1 letter, 3 letters 2 letters, 2 letters 3 letters, 1 letter The entries in the table show the number of letters contained in the first input and the number of letters contained in the second input. For example, "1 letter, 3 letters" means that the two-input combination was made up of the first input being one letter and the second input being three letters. The combination A,C/B/D is such a combination. For each of the six entries in the table, we now need to determine how many distinct combinations can be created. The next table shows this. 1 letter, 1 letter C(4,1)*C(3,1)=12 1 letter, 2 letters
2 letters, 1 letter
C(4,1)*C(3,2)=12 C(4,2)*C(2,1)=12 1 letter, 3 letters 2 letters, 2 letters 3 letters, 1 letter C(4,1)*C(3,3)=4 C(4,2)*C(2,2)=6 C(4,3)*C(1,1)=4 For each case, a product of two combinations is shown. The first factor calculates the number of ways to create the first input for the combination and the second factor shows the number of ways to create the second input for the combination. For example, For a 2-letter, 1-letter combination, the first input, consisting of two letters, can be created in C(4,2) ways, and the second input, consisting of one letter, can be created in C(2,1) ways. Two important points need to be emphasized in these calculations. First, combinations are used based on the fourth entry in the retstrictions list above: When two or three letters comprise an input, they are pressed simultaneously, so the order does not matter. Second, the second factor in each product represents a selection made from the letters remaining after the first input of the combination has been created.
Because the six cases described above are disjoint, we add the individual results to calculate the desired number of combinations possible. This total is 50 combinations.
1.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for parts (a) and (b) of question (1). a. In the expansion of
, what is the value of K for the collected term
?
8!/(4!3!1!) b. How many solutions are there for the equation 1,2,3,4, must be a non-negative integer?
if each
,i=
C(41+4-1,4-1) = C(44,3) 2.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for parts (a) and (b) of question (2). a. A teacher intends to assign letter grades to one class of students according to the following distribution: 3 As, 6 Bs, 7 Cs, and 2 Ds. In how many ways can this grade assignment occur?
a. 18!/(3!6!7!2!) b. Suppose a recurrence relation s(n) is defined as s(n) = s(n-1) - 2·s(n-3) for n larger than 3 with s(1) = 2, s(2) = 5, and s(3) = 1. Determine the absolute difference between the largest and smallest values in the set {s(1), s(2), . . . , s(8)}.
n
s(n)
1
2
2
5
3
1
4
(1) - 2(2) = -3
5 (-3) - 2(5) = -13 6 (-13) - 2(1) = -15 7 (-15) - 2(-3) = -9 8 (-9) - 2(-13) = 17
These values for s(1) through s(8) generate a largest absolute difference of 32, i.e., the absolute difference s(6) = -15 and s(8) = 17. 3.
Here is a portion of Pascal's Triangle.
Consider the sum C(4,0) + C(5,1) + C(6,2) + C(7,3) + C(8,4) = C(9,4). The components of this sum are outlined in the triangle above. a. In the triangle above, outline the following pattern, including the sum represented here with a question mark. Verify that you have included the correct sum. C(2,0) + C(3,1) + C(4,2) + C(5,3) + C(6,4) + C(7,5) = ? What is the sum, expressed as a combination?
See above for outlined pattern. The question mark should be replaced by C(8,5). By straightforward arithmetic one can verify that the outlined values along the diagonal (1, 3, 6, 10, 15, 21) sum to the remaining value (56). b. Determine C(7,0) + C(8,1) + C(9,2) + . . . + C(27,20). Respond using combination notation.
Using the pattern established above, this sum is C(28,20). We note that if the sum is represented by C(a,b), then a is the next positive integer in the sequence as determined by the first numbers in the combinations of the left-side expression. The value b in the sum is determined by the last (largest) second number in the combinations of the left-side expression. c. Determine C(7,7) + C(8,7) + C(9,7) + . . . + C(27,7). Respond using combination notation.
Because C(n,k) = C(n,n-k), we know that the expression here is equivalent to the expression provided in (b) above. Therefore, the sum is the same as the sum conjectured in (b): C(28,20) = C(28,8). BONUS! State and prove the general result that has been illustrated in the problem.
This problem will be left "open" until after the semester exam! 4.
Use this difference table for parts (a) and (b).
a. Complete as many open rows of the table (D1, D2, D3, D4) as necessary to determine the type of equation that will model the relationship between the values in the first two lines of the table (x and f(x)). Write a brief explanation describing how you know what type of equation this will be.
See rows D1 and D2 in table above. With the first constant difference occurring in row D2, we can write a quadratic expression to represent the relationship among the (x,f(x)) pairs that are in the first two rows of the table. b. Use the information in the difference table to generate an explicit formula for the relationship between the values in the first two lines of the table, that is, between x and f(x).
We use the difference table for the general quadratic expression ax^2 + bx + c. x
1
2
3
4
5
6
7
8
(x) a+b+c 4a+2b+c 9a+3b+c 16a+4b+c 25a+5b+c 36a+6b+c 49a+7b+c 64a+8b+c
D1 --- 3a+b > D2 --->
5a+b 2a
7a+b 2a
9a+b 2a
11a+b 2a
15a+b
13a+b 2a
2a
Equating values from this table to the one abovem we compute that a = 2, that b = 1, and that c = -1. The function we seek is f(x) = 2x^2 + x - 1. Here are the first few rows for a triangular table of values. Use this for questions (c) and (d).
c. Write in entries for two more rows of the table. These are rows 4 and 5.
See the additional rows in the table above. d. Use the method of finite differences to determine whether a polynomial exists that models the row sums for this table. If such a polynomial exists, state its degree. If a polynomial cannot be used, state why not. You are NOT required to determine any explicit formula here! Row 1 2
3 4
5
Sum 4 14 32 60 100
D1 ---> 10 18 28 40 D2 ---> 8 10 12 D3 ---> 2 2
Because the first appearance of a constant differrence is in D3, the polynomial in question would be a third-degree polynomial. 5.
Marco's Uncle Oscar gave him two mice, which he named Whiskers and Oscar. But Marco discovered he'd made a big mistake. Oscar should have been named Oscarella! She just had 8 babies, 4 males and 4 females! "Ten mice aren't so many," said Marco to his mother. "They're cute." "Yes," said his mother, "but these cute baby mice will breed when they are 6 weeks old, and continue breeding for a long time. Babies are born 3 weeks after breeding. Each mother mouse will have a litter of 8 babies, half males and half females. Today Marco has 10 mice: Whiskers, Oscarella, and 8 babies. a. How many mice will Marco have in (i) 3 weeks? (ii) 6 weeks? (iii) 9 weeks? (iv) 18 weeks? 0 2 + (4 + 4) weeks
10 = m(0)
3 2+ weeks (4 + 4) + (4 + 4)
18 = m(3)
6 2+ weeks (4 + 4) + (4 + 4) + (4 + 4)
26 = m(6)
9 2+ weeks ((4 + 4) + 4(4 + 4)) + (4 + 4) +
66 = m(9)
(4 + 4) + (4 + 4) 12 2 + weeks ((4 + 4)+ 4(4 + 4) + 4(4+4)) + ((4 + 4) + 4(4 + 4)) + (4 + 4) + (4 + 4) + (4 + 4)
138 = m(12)
15 2 + weeks ((4 + 4) + 4(4 + 4) + 4(4+4) + 4(4 + 4)) + ((4 + 4) + 4(4 + 4) + 4(4 + 4) ) + ((4 + 4) + 4(4 + 4)) + (4 + 4) + (4 + 4) + (4 + 4)
242 = m(15)
18 2 + weeks ((4 + 4) + 4(4 + 4) + 4[4(4+4)] + 4(4+4) + 4(4 + 4) + 4(4 + 4)) + ((4 + 4) + 4(4 + 4)+ 4(4 + 4)+ 4(4 + 4)) + ((4 + 4) + 4(4 + 4) + 4(4 + 4) ) + ((4 + 4) + 4(4 + 4)) + (4 + 4) + (4 + 4) + (4 + 4)
506 = m(18)
b. Write a recurrence relationship for m(n), the number of mice Marco will have n weeks from today. Note that m(0) = 10 and that in part (a) you were asked to determine m(3), m(6), m(9), and m(18).
The desired recurrence relation is m(n) = m(n-3) + 4[m(n-9)]. Today (week n), we have all the mice that existed three weeks ago ( m(n3) ) as well as the baby mice just born from fertile mice that were alive nine meeks ago. There were m(n-9) mice alive nine weeks ago, and every pair produces 4 pairs of babies, so the number of babies today is just 4 times the number that were alive nine weeks ago. 6.
Many professional sports conduct tournaments to finish a season. Some sports have a single-elimination playoff concluding with a sudden-death championship, such as the National Football League's Super Bowl. Other sports, however, end the season with a multi-game series between two teams. For instance, in Major League Baseball, the World Series is a best-of-7 playoff. Two teams compete until one team wins four games. Thus, there may be as few as four games in the playoff or as many as seven. In 1998, the New York Yankees swept the San Diego Padres in four games. From the Yankees perspective, the series record was WWWW, where W represents a Yankee win. In 1997, the Florida Marlins defeated
the Cleveland Indians four games to three. Their 7-game series record was WLWLWLW, where W represents a Florida victory and L a Florida loss. For this type of best-of-7 playoff, how many different won-loss records could there be, from the winner's perspective?
There will be series that last 4 games, series that last 5 games, series that last 6 games, and series that last 7 games. There are some consistent factors across these four cases: (i) The team that wins the series win the final game. Thus, in the sequence of Ws and Ls that comprise a won-loss record, the last element must be a W. In other words, there is only one choice for that position in the sequence. (ii) Prior to the last game, the team that wins the series must have won three games and the team that loses the series must have won whatever other games have been played. So, if a series lasts g games, the team winning the series wins the last game and wins 3 of the (g-1) games prior to that. The losing team wins [(g-1)-3] or g-4 games. For the (g-1) games prior to the final game, there are C(g-1,3) ways to arrange the 3 Ws and the (g-4) Ls that constitute the won-loss record for the g-1 games prior to the final game of the series. With this information we can now count the possible sequences for g = 4,5,6,7 games. If g = 4, there is C(3,3) = 1 way to arrange the 3 Ws prior to the fourth W. That is, WWWW is the only way for a team to win the first four games. If g = 5, there are C(4,3) = 4 ways to arrange 3 Ws and 1 L prior to the fourth W. If g = 6, there are C(5,3) = 10 ways to arrange 3 Ws and 2 Ls prior to the fourth W. If g = 7, there are C(6,3) = 20 ways to arrange 3 Ws and 3 Ls prior to the fourth W. This generates 1 + 4 + 10 + 20 = 35 different won-loss records for this type of series playoff.
1.
a. If we label the rows of Pascal's Triangle starting with n = 0 and the columns starting with k = 0, what is the value of the entry in row 14, column 11?
C(14,11) b. Express P(18,3) in three different ways: (i) as the product of three consecutive integers; (ii) using factorial notation; and (iii) in terms of C(18,3).
(i) 18*17*16 (ii) 18!/15! (iii) C(18,3)*3! c. In how many ways can 12 donuts be distributed to 6 people, allowing that some may receive no donuts or that one person may get them all?
The problem reduces to determining the number of non-negative integer solutions to the equation x1 + x2 + x3 + x4 + x5 + x6 = 12. There are C(12 + 6 - 1,6 - 1) = C(17,5) such ways. d. A biologist plans to complete an experiment with 14 mice, aptly numbered 1 through 14. Six of the mice will receive a double dose of Vitamin E, 5 of the mice will receive a single dose of Vitamin E, and the rest will receive no doses of Vitamin E. In how many ways can this distribution take place?
C(14,6)*C(8,5)*C(3,3) = 14!/(6!5!3!) e. A committee of three is to be formed from among four men (Adam, Bill, Cal, and Dan) and four women (Eve, Fran, Gert, and Hanna). In how many ways can this be done if Adam and Eve refuse to serve on it together?
There are three disjoint cases to consider: I) Neither Adam nor Eve are on the committee: In this case, we must select 3 from the remaining 6 possible committee members: C(6,3). II) Eve is on the committee, Adam is not: We need to select 2 more committee members from the 6 remaining people: C(6,2). III) Eve is not on the committee, Adam is: Same reasoning as for (II): C(6,2). This results in C(6,3) + 2*C(6,2) different ways to create the committee. 2.
The following conjecture is to be proven true by induction or shown to be false using a counterexample: 1 + 2 + . . . + (n-1) + n + (n-1) + . . . + 2 + 1 = n^2.
a. Describe and carry out the first step in the induction process.
We need to show that the result holds for the case n=1. Here, when n = 1, we have 1 = 1^2, which is a true statement. b. Describe and carry out the second step in the induction process.
We assume that the result holds for the case when n = k. for this problem, then, we assume that 1 + 2 + . . . + (k-1) + k + (k-1) + . . . + 2 + 1 = k^2. c. Describe but do not carry out the third step in the induction process.
In this step, we show that, based on the assumption made in Step 2, the result holds for the case when n = k+1. 3.
There are 6 men and 5 women on a committee. A subcommittee of 6 is to be formed. The subcommittee must have no less than two men and two women. In how many ways can such a subcommittee be formed?
We must consider three disjoint cases that represent all possible compositions of the committee. I) There are 2 men and 4 women on the committee. For this case, we choose 2 men from 6 and 4 women from 5. We then match each possible group of men with each possible group of women. This can be done in C(6,2)*C(5,4) ways. I) There are 3 men and 3 women on the committee. For this case, we choose 3 men from 6 and 3 women from 5. We then match each possible group of men with each possible group of women. This can be done in C(6,3)*C(5,3) ways. I) There are 4 men and 2 women on the committee. For this case, we choose 4 men from 6 and 2 women from 5. We then match each possible group of men with each possible group of women. This can be done in C(6,4)*C(5,2) ways. This gives us C(6,2)*C(5,4) + C(6,3)*C(5,3) + C(6,4)*C(5,2) ways to form the committee. 4.
A domino is a rectangle formed by two congruent squares. Each square contains an orderly pattern of "pips" or dots representing a number from zero through six. How many different dominoes can be made under these restraints?
We consider two different cases that are disjoint.
I) The domino has a different number of pips in each of the two squares. In this case, there are 7 numbers possible for one square and 7 for the other. This yields 42 dominoes. Note, however, that the domino 5:6 is not distinguishable from the domino 6:5. Therefore, therefore half of the 42 dominoes counted here will be duplicates. We have 21 different dominoes each with a different pair of numbers. II) The domino has the same number of pips in each of the two squares. There are seven such dominoes, from 0:0 to 6:6. We have a total of 28 dominoes that can be created under the described conditions. 5.
For some positive integer n, how many integers between 0 and 2n inclusive must you pick to be sure that at least one of them is odd?
The set in question is {0,1,2,3,4,...,2n-1,2n}. There are n+1 even numbers in the set, consisting of 0,2,4,...,2n. Thus, it is possible we could pick n+1 values from the original set and still have no even number. The next pick, however, would have to be an odd number. By the pigeonhole principle, we need n+2 values to assure that at least one of them is odd. 6.
Suppose that Jay Cool and the Gingos are playing at the Starlite Theatre. The theatre has one section of seats, arranged with 70 seats in the first (front) row, 72 seats in the second row, 74 seats in the third row, and so on, for a total of 30 rows. The seats are numbered from left to right, with the first seat in the first row being #1, the first seat in the second row #71, and so on. a. Write a recurrence relation rel ation and an explicit formula for R(n), the number of seats seat s in Row n. Be sure to include information about initial conditions. Use your results to determine the number of seats in the last (30th) row.
Recurrence Relation: Each successive row contains 2 more seats than the previous row. So if we begin with R(1) = 70 as the initial condition, we get R(n) = R(n-1) + 2 as the recurrence relation. Explicit Formula: This is a linear pattern, noting the constant difference of 2 seats between each consecutive pair of rows. The explicit formula is R(n) = 68 + 2n, where n is the row number and n spans the positive integers from 1 through 30. Either relationship yields the value R(30) = 128 seats. b. Write either an explicit formula or a recurrence relation for S(n), the sum of all
seats through Row n. Use that to determine the total number of seats in the theatre.
Recurrence Relation: We begin with S(1) = 70 as the initial condition. Each successive sum will contain all the seats prior to that row plus all the seats in that row. The number of seats in any particular row was determined above: R(n) = 68 + 2n. This leads to S(n) = S(n-1) + (68 + 2n) as a recurrence relation for the row sums. Explicit Formula: This is a quadratic pattern, determined using a difference table and observing that the first constant difference occurs in the second differences, that is, in line D2 of the table. Associating this specific difference table with the difference table for the general quadratic results in an explicit formula of S(n) = n^2 + 69n. Either relationship yields the value S(30) = 2970 seats. 7.
The set of letters {A,A,A,B,B,B,B,C,C} is to be used to create k-element subsets. a. State the range of values for k that is possible for this situation.
Allowing for a subset that is empty as well as a subset containing all elements for the set, k could be any value in the set {0,1,2,...,9}. b. How many subsets subsets in all can be created?
If we allow for duplicate letters to be used as appear in the original set (i.e., one A is different from another) there are 2^9 subsets. If we do not allow for this, for example, only one subset exists of the form {A,B,C,C} no matter which of the As and Bs are chosen, then there are 60 unique subsets. This can be determined by brute force listing of all unique subsets and can also be determined by summing the coefficients of an appropriate generating function.
Create a generating function that can be used to determine the number of 6&endash;element subsets that will have at least one A and at least two Bs. c. Show the factors of the generating function before expanding them. Explain what each factor represents within the context of this problem.
The factors are (x+x^2+x^3)(x^2+x^3+x^4)(1+x+x^2). Concentrate on the exponents in the terms of each factor. The first factor represents the possible number of As we could include: 1,2, or 3. The second factor represents the possible number of Bs to include: 2,3, or
4. The last factor represents the possible number of Cs we could include: 0, 1, or 2. d. Expand your response to part (c).
x^3 + 3x^4 + 6x^5 + 7x^6 + 6x^7 + 3x^8 + x^9 e. Use part (d) to determine the number of 6-element subsets that will have at least one A and at least two Bs. Include a brief explanation for how you used part (d).
According to the expansion in (d), there are 7 6-element subsets of {A,A,A,B,B,B,B,C,C} that include at least one A and at least two Bs. This value is the coefficient of the term in the expansion for which x is raised to the 6th power, where the 6 represents the number of elements in the subset. 8.
Determine numerical values for x, y, and z that make expansion of
a collected term in the
.
We must have x+y+z=k and we must have k!/(x!y!z!)=210. By trial and error, one set of values that satisfies these conditions is k=7 and x,y,z matched with elements of the set {2,2,3}. Can you come up with other sets, or show that this is the only solution? 9.
a. How many positive integers are there that are composed of n digits such that each digit is 1, 2, or 3?
For an n-digit number, each digit could be one of three values, either 1, 2, or 3. This means there are 3^n positive integers that meet the desired condition. b. Of these, how many contain all three of the digits 1, 2, and 3 at least once?
We apply the inclusion/exclusion principle here to eliminate the positive integers counted above that DO NOT contain each of 1, 2, and 3 at least once. We consider ways to create n-digit numbers that are missing at least one of 1, 2, and 3. Begin by answering the following question: How many ways are there to create an n-digit number composed of no more than two of the values 1, 2, or 3? Each of the n values in such a number can be selected from only two choices, so there are 2^n such numbers. Thus, if we want an n-digit number with only 1s or 2s in it (i.e., 3 is not included), there are 2^n ways to create it. This would be true for an n-digit number with only
1s or 3s in it (i.e., 2 is not included), as well as for an n-digit number with only 2s or 3s in it (i.e., 1 is not included). Therefore, there are 3*2^n n-digit numbers, from the set determined in (a) above, that have at most two different digits in them. But this collection of n-digit numbers includes some duplication, for some of the n-digit numbers have only one unique digit in them, and we have counted those twice. So we must add back the number of ndigit positive integers that contain only one unique digit from the set {1,2,3}. There are just three such numbers, an n-digit number composed of all 1s, an n-digit number composed of all 2s, and an ndigit number composed of all 3s. So, by the inclusion/exclusion principle, we have 3^n - 3*2^n + 3 ndigit numbers that contain each of the three digits 1,2, and 3, at least once. 10.
My nephew Seth noticed that Kellogg's cereals offers a set of 5 cartoon characters in its current cereal selections. One cartoon character is in each specially marked box of cereal and the cartoon characters are equally distributed among the cereal boxes currently coming off the production line. Seth has been around me enough so that he can figure out some probabilities. He knows that the probability of getting a complete set by purchasing less than five boxes is 0. He also explained how to determine the probability he could get a complete set by buying exactly 5 boxes of cereal. He said that there are 5! ways he could buy 5 boxes and get all different characters. He also told me that there were 5^5 different ways to get a set of 5 characters, not necessarily all different. The probability of getting a complete set in the first five purchases is 5!/(5^5). Here's where he needs your help: What's the probability that it takes exactly 6 boxes of cereal to collect a complete set of 5 cartoon characters?
For it to require exactly 6 boxes to assemble a complete set of 5 cartoon characters, two conditions must hold. •
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Among the first 5 boxes purchased, there must be exactly 4 different cartoon characters. This means there are two of one type and one each of three other types. The sixth box purchased must contain the missing, or fifth, character.
We determine in how many ways each of these conditions can occur. The second condition is straightforward in determination: There are 5 different characters that could occupy the last position in the line up,
that is, be in the sixth box. For each of those 5 possibilities in the last position, we now determine ways to meet the first condition. Two things need to be considered: (a) Which or the four remaining letters will be repeated among the first five boxes opened? (b) What is the arrangement of the five letters (four unique letters) once (a) has been determined? For (a), there are 4 choices, because one letter has already been used as the last (sixth box). For (b), a way to think of this is to determine the number of arrangements of a 5-letter word composed of 4 different letters, such as 5-letter words composed on 4 letters from the set {A,B,C,D}. One such word is ABCDA, and another is DCBBA. This is a problem we've solved several times. We know there are 5!/(2!1!1!1!) arrangements of those letters. Associating the results from considering these conditions, we have 5*4*[5!/(2!1!1!1!)] or 1200 such arrangements of the cartoon characters. In all, there are 5^6 different ways to get cartoon characters in the first six boxes, for with each box there are 5 items than could be in it. Our desired probability is 5*4*[5!/(2!1!1!1!)]/(5^6) or 1200/(5^6), approximately 7.65%. This is just less than double the probability Seth determined for the first 5 boxes, which was 3.84%. BONUS!
Each week two World Professional Tennis Organizations determine who are the #1 players in the world for both womens' and mens' professional tennis. a. Steffi Graf was #1 in womens' tennis without interruption over the period Monday, August 17, 1987, until Sunday, March 10, 1991. This is the greatest number of consecutive weeks that any woman or man has been #1 in the professional rankings. How many consecutive weeks were there in Steffi Graf's record? Clearly show your work in solving the problem. b.Steffi Graf was also ranked #1 during these time periods: • • • • •
August 5 to August 11, 1991 August 19 to September 8, 1991 June 7, 1993, to February 5, 1995 February 20 to February 26, 1995 April 10 to April 30, 1995
Including your response to (a), for how many total weeks of her career, through 30 April 1995, was Steffi Graf ranked #1?
The solution to this problem will appear soon. Keep trying!
1.
A mathematics educator is planning to survey various professionals. Surveys will be look at sent to 44 biologists, 29 chemists, 37 physicists, 51 endocrinologists, and 22 possible epidemiologists. How many responses must the mathematics educator receive in order solution to guarantee that there will be at least 20 responses from the same professional group?
2.
Frank intends to invest his life savings of $200,000 in five mutual funds from a list of 20 such funds prepared by his mother, an investment banker. (a) How many different investments are possible if Frank invests $40,000 in each fund?
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(b) How many different investments are possible if Frank invests $20,000 in each of two funds, $40,000 in a third, $60,000 in each of two remaining funds? 3.
Six men and seven women, all of different heights, stand in a waiting line at the bookstore. (a) How many arrangements of these people are possible if the men stand in succession?
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(b) How many arrangements of these people are possible if two women stand at the very front of the line and two men are at the very end of the line? (c) How many arrangements of these people are possible if the men and women must alternate positions within the line? 4. One definition for the word abracadabra is "a charm or incantation." How many unique arrangements are there for the letters in the word abracadabra? 5.
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Two rooks on a chessboard attack each other if they are on the same row or the same column. Determine the number of ways in which 8 non-attacking identical rooks can be placed on the chessboard shown here. Squares with an "X" indicate forbidden positions that cannot be occupied. look at possible solution
6.
To complete this problem: Do ONE of the two problems listed under (I) below, OR
6-I-a: look at possible solution 6-I-b:
Do BOTH problems listed under (II) below. I: (a) Prove that C(1,1) + C(2,1) + ... + C(n,1) = C(n + 1,2) for every positive integer n. (b) Prove that 1(1!) + 2(2!) + ... + n(n!) = (n + 1)! - 1 for every positive integer n. II: (a) Prove that r*C(n,r) = n*C(n - 1,r - 1) for n >= r >= 1. (b) Prove that C(n,m)*C(m,k) = C(n,k)*C(n - k,m - k) for k <= m <= n.
1. 96 responses This problem may be solved by applying the Pigeonhole Principle. Imagine a basket for each of the five professional groups, into which completed surveys are deposited as they are received. The worst possible situation that fails to meet the condition is that each of the five baskets will have 19 replies. This represents 95 completed surveys. Upon receiving the 96th reply, however, we can be sure that at least one basket contains at least 20 replies. 2. (a) C(20,5) Here, order or arrangement is not important because each of the 5 funds is to receive $40,000. There are C(20,5) sets of 5 funds that can be selected for investment. (b) P(20,5)/(2!2!) Now the match-up of dollars with a particular fund does matter, because differing amounts are being invested. There is duplication, however, in that two funds receive $20,000 and two others receive $60,000. For example, if stocks A,B,C,D,E are chosen, $20,000 in A and B, $40,000 in C, and $60,000 in D and E is the same as $20,000 in B and A, $40,000 in C, and $60,000 in D and E. There are P(20,5) ways to arrange 5 of the 20 funds, and we divide by 2!2! to account for duplication of amounts invested. 3. We assume that all individuals are distinguishable, because we are told that they all differ in height. (a) 8!6! Chunk the men into one "position" in line. This reduces the task to placing 8 objects in line (7 women and 1 chunk of men). This can be done in 8! ways. The six men can be arranged in 6! ways within their group. (b) P(7,2)*P(6,2)*9! There are P(7,2) ways to arrange two women at the front of the line. Likewise, there
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are P(6,2) ways to arrange two men at the end of the line. This leaves 9 people to fill the remaining positions, which they can do in 9! ways. (c) 7!*6! In order to alternate, the line must form with a woman in front and then alternate from there. This means each position in line is assigned to a gender in one and only one way. Within that restriction, there are 7! ways to arrange the women and 6! ways to arrange the men. 4. 11!/(5!2!2!) There are 11 letters in the word so there are 11! ways to arrange the letters. With identical letters a, b, and r, however, some of those 11! arrangements are duplicates. There are 5! ways to interchange the letters a, 2! ways to interchange the letters b, and 2! ways to interchange the letters r. We divide by these values to account for the duplication. 5. 3*7! Label the rows and columns to allow for convenient discussion, calling the column c1 through c8 left to right and the rows r1 through r8 top to bottom. We first place a rok in r1 because it is the most restrictive. There are three places for a rook in r1. Once a rook is placed, its row and column are eliminated because the rooks must be arranged in non-attacking positions. Without loss of generality, suppose the rook is in c8 of r1. We now move to r2 where there are seven spots available for a rook. Suppose the rook is placed in c1 of r2. This eliminates c1 and r2 for further placement. Continue this progression of placing a rook in available spots and eliminating that row and column from further placement. Eventually, one spot remains in r8 for the last rook. This gives us 3·7·6· 5·4· 3·2·1 ways to place the eight rooks into non-attacking positions. 6. I. Prove (a) or (b) using induction. (a) Prove that C(1,1) + C(2,1) + ... + C(n,1) = C(n + 1,2) for every positive integer n. Step 1: Show the result holds for n = 1: Is C(1,1) equal to C(1 + 1,2)? Yes, because C(1,1) = 1 and C(1 + 1,2) = C(2,2) = 1. Step 2: Assume the result holds for n = k: Here, we assume that . Step 3: Use the assumption from Step 2 to prove that the result holds for n = k + 1. That is, we seek to show that . From the assumption in Step 2, add C(k + 1,1) to each side: .
The LS of this equation matches the LS of the equation we seek to justify as correct. Now, simplify the RS:
This last expression is just the result we seek, as shown in the initial equation of Step 3. We have proved through mathematical induction that C(1,1) + C(2,1) + ... + C(n,1) = C(n + 1,2) for every positive integer n. (b) Prove that 1(1!) + 2(2!) + ... + n(n!) = (n + 1)! - 1 for every positive integer n. Step 1: Show the result holds for n = 1: Is 1(1!) = (1 + 1)! - 1? Yes, because 1(1!) = 1 and (1 + 1)! - 1 = 2! - 1 = 1. Step 2: Assume the result holds for n = k: Here, we assume that . Step 3: Use the assumption from Step 2 to prove that the result holds for n = k + 1. That is, we seek to show that . From the assumption in Step 2, add(k + 1)(k + 1)! to each side:
The LS of this equation matches the LS of the equation we seek to justify as correct. Now, simplify the RS:
This last expression is just the result we seek, as shown in the initial equation of Step 3. We have proved through mathematical induction that 1(1!) + 2(2!) + ... + n(n!) = (n + 1)! - 1 for every positive integer n.
II. Use your knowledge of combinatorial notation and algebra to prove that the leftand right-side expressions of each equation are equivalent. (a) Prove that r*C(n,r) = n*C(n &endash; 1,r &endash; 1) for n <= r <= 1. Simplify the LS: Simplify the RS: The two expressions are equivalent, thereby justifying the result. (b) Prove that C(n,m)*C(m,k) = C(n,k)*C(n &endash; k,m &endash; k) for k <= m <= n. Simplify the LS: Simplify the RS: The two expressions are equivalent, thereby justifying the result. 1. Respond to each of these questions by placing your solution in the blank. You do not need to generate explanations for questions (a) through (e).
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a.Express C(20,6) in terms of P(20,6). b.How many distinct arrangements exist for the letters in the word TETRAMETER? c.In the expansion of (w + x + y + z)^19, determine the value of the coefficient K in the collected term Kw^2x^9y^5z^3. d.Express C(10,5) - C(9,5) as a single combination. e.Express in simplest form the sum that includes every other term, starting with the first term, in the (ordered) 20th row of Pascal's Triangle. 2. How many ways can the letters in the word oblique be arranged so that no look at possible two consonants are adjacent? solution 3. Write the collected terms in the expansion of (w - 2k)^4. Simplify the numerical coefficients.
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4. A flag is to be created for the new country of Slynskonskia. The country's look at possible founders agreed that: solution
•
•
the flag should be composed of three solid-colored horizontal stripes of identical width, no two adjacent stripes could be the same color, and
•
there would be no designated "top" of "bottom" to the flag.
If there are five colors to choose from, how many different flags could be made for Slynskonskia under these conditions? 5. A city ballot includes 12 referendum questions. For each question, voters choose YES or NO to express their preference.
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a.How many different marked ballots could be submitted, given that a choice of YES or NO is made for each of the 12 questions? b.How many different marked ballots could show 5 questions marked YES and 7 questions marked NO, given that a choice of YES or NO is made for each of the 12 questions? c.If voters may abstain from choosing a response on any and all questions, how many different ballots could be submitted by voters? 6. A game of super-dominoes is played with pieces divided into three cells instead of the usual two. The set contains all possible triplets of values from triple blank to triple six, with no duplications. For example, the set does not include both 2-3-1 and 1-3-2 because these are simply reversals of each other. The set does contain 3-1-2, however.
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How many pieces are in the set?
Test #2: Possible Solutions
1. (a) P(20,6)/6! (b) 10!/(3!3!2!) (c) 19!/(2!9!5!3!) (d) C(9,4) (e) 2^19 2. 4!*P(5,3) Place the vowels first. There are 4! ways to do this. There are now five places among the vowels into which consonants may be placed so as to keep them from each other. Arrange the 3 consonants among the 5 available spots, done in P(5,3) ways. 3. Here is the expansion.
4. 50 Consider two cases depending on the number of different colors in a flag. Case I: Flags with three different colors There are P(5,3) ways to create these flags, but because there is no distinction between the top and bottom of a flag, we have twice as many flags as exist. We divide by 2 to account for flags of, say Red-Green-Blue looking the same as Blue GreenRed. This gives us 30 different flags. Case II: Flags with two different colors The only way to have two colors is for the outside stripes to be the same color and the middle color different from the outside. There are P(5,2) ways to create this type of flag. No division by 2 is necessary because of the symmetry of the colors. Red-GreenRed has only been counted once, for example. This yields 20 more flags. We add the results because Case I and Case II represent two non-overlapping sets: One flag cannot have two different colors and three different colors. 5. (a) 2^12 For each of the 12 questions there are two choices, YES or NO. We use the multiplication principle because each choice is, we assume, independent from the rest. (b) C(12,5) = C(12,7) From among the 12 questions, choose 5 for YES or, equivalently, choose 7 for NO. There is no accounting for the arrangement here, for we are simply gathering 5 (or 7) questions. (c) 3^12 For each of the 12 questions there are now three choices, YES or NO or ABSTAIN. We use the multiplication principle because each choice is, we assume, independent from the rest. 6. 196 dominoes There are several ways to solve the problem. One approach is to count the dominoes that can be created for each of three cases, depending on how many different numbers are on the domino, and then add the results. Case I: Dominoes that show exactly one unique number (7 dominoes) There is one and only one way to arrange one number three times on a domino. There are seven numbers to choose from, but once a choice is made for one of the spots on a domino, the other two are determined. There are, then, P(7,1) = 7 such dominoes. Case II: Dominoes that show exactly two different numbers (84 dominoes) There are two different ways to have two different numbers appear. (i) Outside values equal, middle value different, such as 3-6-3 or 2-0-2. There are P(7,2) = 42 of these, formed by choosing one of 7 values for the middle spot and one of 6 for the other two. There is no duplication here, because we've only filled two spots; the third is determined by one of the first two.
(ii) Two equal values next to each other, such as 1-1-2 or 4-5-5. We count only dominoes of the type x-x-y here, for the domino y-x-x is the same. There are 7 ways to choose the value that appears twice on this dominoe and 6 ways to choose the single value. There are P(7,2) = 42 such dominoes. Case III: Dominoes that show exactly three different numbers (105 dominoes) These are P(7,3) = 210 of these, counted by choosing from among 7 values for the first spot, from among 6 for the second, and from among the remaining 5 for the lasy spot. However, this technique counts the same domino twice, for the domino 3-2-6, for instance, is also counted as 6-2-3, but this is actually only one domino. We divide P(7,3) by 2 to get the actual number of different dominoes of this type. There are 105 of these. 1. Consider the expansion of the multinomial (m + n + p + s + v)^11. a) Determine the number of uncollected terms in the expansion of this multinomial.
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b) Determine the number of collected terms in the expansion of this multinomial. c) Determine the value of K for the collected term Km^4n^2pv^4 BONUS! A collected term in the expansion of the multinomial shown above has coefficient 11. How many collected terms will have that coefficient? 2. Consider the word microfibrillar. a) How many unique arrangements are there for the letters in this word? b) How many ways are there to arrange the letters in the word and keep the vowels from being adjacent to one another? BONUS! Write down a legitimate English-language word that fits all the following criteria. (Webster's Unabridged, in STV 313, will be used to certify words): i) The word has no fewer that 6 letters. ii) At least one of the letters appears more than once. iii) There are no less that 16,480 unique arrangements of the letters in your word. Justify your response. DOUBLE BONUS!! Determine the maximum number of letters for a word that meets all the following conditions: i) There are at least 4 unique letters in the word. ii) At least 2 different letters in the word repeat.
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iii) There are fewer than 8910 unique arrangements of the letters in the word. Justify your response. look at 3. Joannie Jorgensen volunteers at the downtown Senior Citizen Center. She's responsible for keeping records and reporting statistics that describe possible solution each day's luncheon clientele. She records the following information for each diner:
• the gender of the diner • whether the diner is a local resident or from out of town • whether the diner is a member of AARP • whether the diner has previously eaten at the t he Center Last Friday, Joannie reported information about the day's luncheon diners. Here's the beginning of her report: DINER REPORT FOR FRIDAY filed by Joannie Jorgensen 57 diners in all: 17 male/local resident/AARP members/returning diners 16 female/local resident/AARP members/returning diners 0 male/local resident/AARP members/first-time diners 3 female/local resident/AARP members/first-time diners . . . and so on . . . a) How many different descriptions are possible for diners according to the records kept by Joannie? b) Yesterday there were 38 diners. How many different different reports, similar to the one above, could have been filed by Joannie yesterday, given that some descriptions may have had no diners, such as the third entry in the example? 4.
Two octopi took part in a friendly tentacle-to-tentacle wrestling match. Each managed to pin 4 of its opponent's tentacles with 4 of its i ts own. In how many ways could the match have taken place?
5. 5. Fifty tickets, numbered consecutively consecutively from 1 to 50, are placed in a #2 mayonnaise jar on Funk & Wagnall's porch. Two tickets are drawn from the jar. The order the two tickets are drawn is not important. a) Verify that there are 1225 ways for this to occur. Explain your verification. b) Of the 1225 possible possible pairs of tickets, how many pairs show two numbers whose difference is 10 or less?
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6. Here is a problem situation and the work submitted by a student: A shelf is to contain nine different books, six different paperback books and three different hardback books. books. If the paperback books must be shelved in pairs pairs (that is, exactly two paperback books must be adjacent to each other), how many ways can the nine books be shelved? First, place the hardback books so 2 paperbacks can go between them. There are 3! ways this can be done.
To complete the arrangement, there are 8 places left to place 6 paperback books. Since the books are different and order matters, we have a permutation, P(8,6). The number of arrangements of the two types of books must be multiplied together since the total arrangement depends on the placement of each type. Solution: 3!*P(8,6) Analyze the student's work. Is it correct? If so, identify the key step or steps in the student's solution. If the work is incorrect, identify the error or errors in the student's work and suggest what should have been done.
Test #3: Possible Solutions
1. a) 5^11 To expand the multinomial, you make a choice of one term from five available availa ble in each of the 11 factors (m+n+p+s+v). Because the choices are independent, we use the multiplication principle: 5*5*5*5*5*5*5*5*5*5*5 5*5*5*5*5*5*5*5*5*5*5 b) C(15,4) Each collected term of the expansion is of the form where M+N+P+S+V=11. Thus we seek the number of nonnegative-integer solutions to the equation M+N+P+S+V=11. There are C(11+5-1,5-1) ways to solve this equation. c)11!/(4!2!4!) There are 11! arrangements of the letters mmmmnnpvvvv. We divide by (4!2!4!) to account for the duplicate letters m, n, and v. BONUS Solution: 20 collected terms For the coefficient K of the collected term to be 11, we must have means the collected terms look like
. This
. We need to determine the number of ways
to arrange two elements from {m,n,p,s,v} so that one is x1 and the other is x2. This is P(5,2)=5*4. 2. a) There are 14! arrangements of the letters in the word. We divide by (3!3!2!) to account for the duplicate letters. b) Place the consonants first. There are 9!/(3!2!) 9 !/(3!2!) ways to do this. This leaves us with 10 spots into which we place 5 vowels. This can be done in P(10,5) ways. We divide by 3! to account for the letter i appearing 3 times. Bonus Solution: Many solutions are possible. The word you submitted was checked to determine whether all three criteria cri teria were met. Double Bonus Solution: 13 letters Consider some examples. There must be at least 6 letters in the word, based on conditions (i) and (ii). The word picnic is such a word. We symbolize the unique and total number of letters as {1,1,2,2} to show there are 4 unqiue letters (elements in the set) and 6 letters in all (sum of the values in the set). set ). There are 6!/(2!2!)=180 unique arrangements of the letters in this word. If we continue to duplicate letters and introduce no new letters, the progression of unique arrangements is (using ( using the symbolism shown above) {1,2,2,2}--> 630 arrangements, {2,2,2,2}--> 2520 arrangements, {2,2,2,3}--> 7560 arrangements, and {2,2,3,3}--> 25200 arrangements. This shows that we could have at most 9 letters. Another way to duplicate letters and not increase the number of new letters is to only further duplicate one letter. Here's what we get for that scenario: {1,1,2,3}--> 420 arrangements, {1,1,2,4}--> 840 arrangements, {1,1,2,5}--> 1512 arrangements, {1,1,2,6}--> 2520 arrangements, {1,1,2,7}--> 3960 arrangements, {1,1,2,8}--> 5940 arrangements, {1,1,2,9}--> 8580 arrangements, and {1,1,2,10}--> 12012 arrangements. This shows that we could have at most 13 letters. Yet another option is to simply add additional new letters without duplicating any. Here's what happens: {1,1,1,2,2}--> 1260 arrangements and {1,1,1,1,2,2}--> 10080 arrangements. This shows that we could have at most 7 letters. There are no other ways to uniquely increase the number n umber of letters, duplicate or otherwise. 3. a) 2^4=16 For each of 4 categories there are 2 choices. b) C(53,15) There are 16 categories to fill fil l with the information from the 38 diners, and it's
possible that a category may have zero entries. We're determining the number of nonnegative-integer solutions to the equation x1+x2+...+x16=38. This is C(53+161,16-1). 4. Consider the stages of the process of selecting and matching tenacles between the two critters. One way to consider this is to first choose 4 tenacles from Octopus A. this can be done in C(8,4) ways. Now choose 4 tenacles from Octopus B, done in C(8,4) ways. Now permute the 4 chosen tenacles of Octopus B and match each permutation with the 4 already selected from Octopus A. This last step can be done in P(4,4)=4! ways. Let this be the situation where Octopus A pins 4 of Octopus B's tenacles. While this is going on, each octopus has 4 tenacles remaining. Permute the idle 4 tenacles of either octopus (but not both), done in P(4,4)=4! ways. Match these with the remaining 4 tenacles of the other octopus. Because each of these steps is independent from the others, we multiply the results of each step to get C(8,4)*C(8,4)*P(4,4)*P(4,4). This is equivalent to P(8,4)*P(8,4), and you can make a combinatorial argument to satisfy that form as well. 5. a) C(50,2)=1225 The value 1225 is just C(50,2) which precisely matches the situation presented in the problem. We're simply grabbing two unique objects from the 50 that are in the jar, and the order we pull them from the jar is not important. b) 445 pairs This solution requires some degree of enumeration for the several cases that exist. We need to consider what could be the difference between any pair drawn and then count those that meet the condition of being within 10 of each other. Concentrate on one number in a pair. If it ranges from 11 to 40, it has 20 different values (10 less than it and 10 greater than it) that it could pair with and yield a difference 10 or less. If one number is 10 or 41, it has 19 different values it could pair with and generate a difference of 10 or less. If one number is 11 or 42, is has 18 different values, and so on, to one number being 1 or 50, pairing with only 10 possible values. We now add the possibilities: 30(20)+2(19)+2(18)+...+2(10). This sum is 890. We must divide that be 2, however, because, we are not distinguishing a first number from a second in any pair. For instance, pairing 14 with 9 is counted twice, first when looking at 14 matching with 20 possible values and then 9 matching with 18 possible values. Therefore, our correct solution is 445 pairs. 6. First consider a legitimate solution to the problem and then compare it to that of the student. We use the hardback books (HBB) as barriers to the paperback books (PBB), so place the HBB first, in 3! ways. This gives us 4 places to places pairs of PPB. Select 3 of
those, in C(4,3) ways. Now, consider how the PBB can be rearranged in their chosen places. Despite the fact that the PBB are paired, there are P(6,6)=6! ways to arrange them in their places. Each of the three steps in the process are independent of one another, so there are 3!*C(4,3)*6! ways to carry out this shelving task. Now consider the student's solution. Comparing it to mine, the first step seems legitimate: placing the HBB as barriers. The next step, however, is incorrect. The student's arrangement of 6 books among the 8 spots designated does not assure pairing of PBB, despite the solver's drawing that shows "pairs" of available spots between the HBB. The student's solution does nothing to prevent a single PBB to appear somewhere between two HBB. BONUS QUESTIONS Correct solutions were not presented for all three problems, so the bonus solutions will be left open. 1. Every weekday at lunch, Felix eats one apple and only one apple. Felix buys his apples from a grocer who always has the same 8 varieties of apples available. (a) If Felix purchases apples only from this grocer and he wants to have a different variety of apple each weekday, Monday through Friday, how many different "apple menus" can Felix create? (b) What if Felix decides it's okay to have the same variety of apple more than once during a week (M-F)? Now how many "apple menus" can he create? 2. Behind the scenes at the upcoming 1996 Westminster Kennel Club Show in Madison Square Garden, there will be a room that contains 6 English Cocker Spaniels and 12 Fox Terriers. In how many ways can the 6 Cockers be paired with 6 of the Terriers? 3. A bag contains 26 red tokens labeled a, b, . . . , z and 42 green tokens labeled 2, 4, . . . , 84. (a) In how many ways can 20 tokens be selected from the bag? (b) In how many ways can 20 tokens of one color be selected from the bag? 4. Let us define a "word" as any arrangement of one or more letters of the 26-letter alphabet. How many two-letter "words" can be formed if each word must have distinct letters and no word can contain two consecutive letters of the alphabet? For example, bb is not a word, nor are jk and kj. 5. A group of 24 married couples has gathered at a northern Minnesota smelt fry. How many individuals must be chosen from the group to ensure that at least two of the persons chosen are a married couple? 6. Show that for any six positive integers, there must be at least one pair whose absolute difference is divisible by 5. Your justification should include more than simply an example that illustrates the result.
1. Express P(14,4) in three different ways:
(a) as the product of four consecutive integers; (b) using factorial notation; and (c) in terms of C(14,4). 2. How many distinct arrangements exist for the letters in the word PREFERRER? 3. If we label the rows of Pascal's Triangle starting with n = 0 and the columns
starting with k = 0, what is the value of the entry in row 20, column 3: (a) expressed using combination notation; and (b) expressed as a base-10 number. 4. Replace w, x, y, and z in C(18,9) = C( w,x) + C( y,z) to illustrate Pascal's Formula, a
fundamental relationship that exists in Pascal's Triangle.
5. Juanita is a political satirist. She claims to know enough jokes today so that she
could tell a different set of three jokes in her warm-up act, every night of the year, for at least 40 years. What is the minimum number of jokes she must know? NOTE: The set of jokes {A,B,C} is considered one set of jokes, no matter what order Juanita tells the three jokes. 6. Here is a problem situation and a proposed solution. Problem: How many di f f er ent t hr ee- di gi t br i ef case- l ock arr angement s
can be cr eat ed i f an ar r angement cont ai ns t hr ee di st i nct di gi t s wi t h no mor e t han one di gi t gr eat er t han 7? Solution: Thi nk of
f i l l i ng t hese t hr ee spaces wi t h di gi t s: __ ___ _ ___ _ __ ___
As CASE I , suppose t he ar r angement has no di gi t s gr eat er t han 7. We have 8 di gi t s t o choose f r om f or t he f i r st posi t i on, 7 f or t he second, and 6 f or t he t hi r d. We mul t i pl y t o get 8*7*6 ar r angement s wi t h no di gi t s gr eat er t han 7. For CASE I I , suppose t he ar r angement has one di gi t gr eat er t han 7. We have 2 di gi t s t o choose f r om f or t he di gi t gr eat er t han 7, 8 t o choose f r om f or t he next di gi t , and 7 t o chose f r om f or t he f i nal di gi t . We mul t i pl y t o get 2*8*7 ar r angement s wi t h one di gi t gr eat er t han 7. The t wo cases ar e di sj oi nt , f or no ar r angement can have no di gi t s gr eat er t han 7 and one di gi t gr eat er t han 7 at t he same t i me. Ther ef or e, we add t he r esul t s of t he t wo cases. Ther e ar e 8*7* 6 + 2*8*7 poss i bl e ar r angement s.
Comment on the proposed solution by verifying or disputing the argument. 7. Consider the expansion of (2 a + b)^5.
(a) How many collected terms are there? (b) Write out the first three collected terms, beginning with the term containing the factor a^5. 8. Consider the expansion of ( e + f + g + h)^25.
(a) How many uncollected terms are there? (b) How many collected terms are there? (c) What is the coefficient of the collected term that contains the factor ef ^6g^10h^8
9. George works 16 blocks west and 6 blocks south of his home. All streets from his
home to his workplace are laid out in a rectangular grid, and all of them are available for walking. On his walk to work, George always stops at Buddie's Bakery & Deli, located 4 blocks west and 2 blocks south of his home. On his walk home from work, George always stops at Brink's Bank, located 3 blocks north and 4 blocks east of her workplace. If he walks 22 blocks from home to work and 22 blocks from work to home, how many different round-trip paths are possible for George? 10. A shelf is to contain 16 books. There are 11 red books and 5 green books. If no
two green books can be adjacent to each other, and the books are distinguishable only by color, how many ways can the 16 books be shelved? Possible Solutions: Test #2 1a) 14€13€12€11 1b) 14!/10! 1c) C(14,4)€4! 2) 9!/(1!4!3!1!) There are 9! ways to arrange the 9 letters in the word. We divide by (1!4!3!1!) to account for the arrangements of identical letters that result in no visible rearrangement of the letters. 3a) C(20,3) 3b) 1140 4) C(18,9)=C(17,9)+C(17,8) Pascal's Formula states that C(n+1,k+1)=C(n,k+1)+C(n,k). 5) 46 jokes If Juanita works every night for the next 40 years, there are at least 14610 nights, allowing for leap year every four years (40€365+10). We then need to determine a value for n so that C(n,3)„14610. By trial and error or by solving the inequality n!/[3!(n-3)!]>=14610, over the positive integers, we determine that n„46 satisfies the desired relationship. 6) Case I of the proposed solution is accurate. For Case II, the solution as presented uses an incorrect assumption that the left-most digit will always be the digit that is greater than 7. This is more restrictive than stated in the original problem. To rectify this, we should consider the two other possible positions for the digit greater than 7. In each case, the argument as stated will show there are 2€8€7 possible arrangements. So for Case II, there are three times as many possibilities as suggested by the solver. Therefore, there are 8€7€6 + 3(2€8€7) possible arrangements for the briefcase lock. 7a) 6 In the expansion, the variable a will appear as a factor from 0 to 5 times in any one
term. 7b) 32a^5 - 80a^4b + 80a^3b^2 This is simplified from C(5,0)(2a)^5(-b)^0 + C(5,1)(2a)^4(-b)^1 + C(5,2)(2a)^3( b)^2. 8a) 4^25 For each of 25 factors (e + f + g + h), there are four terms to choose from to carry out the expansion. 8b) C(28,3) Each collected term must be of the form [25!/(E!F!G!H!)]e^E€f^F€g^G€h^H, where E+F+G+H=25 solved over the nonnegative integers. There are C(25+4-1,4-1) solutions to this equation. 8c) 25!/(1!6!10!8!) Refer to the general representation for a collected term, shown in (8b). 9) [C(6,4)€C(16,12)]€[C(7,4)€C(15,12)] George must pass through the bakery. There are C(6,4) different 6-block routes he can take. He then must go 16 blocks to work, including 12 westerly blocks. There are C(16,12) routes from the bakery to work. Using the same reasoning for the return trip, there are C(7,4) routes to the bank from work and C(15,12) routes from the bank to George's home. Because each route to work can be matched with each route from work, we multiply to get the total number of round-trip paths George may take. 10) C(12,5) Line up the red books first. There are 12 spaces among the red books, into which we must place the 5 green books. This can be done in C(12,5) ways. We use combinations rather than permutations because we do not concern ourselves with either how the 11 red books nor the 5 green books are arranged among themselves. The books are distinguishable only by color. 1. Consider the expansion of (a + b)^10. (a) Determine the coefficient K of the collected term Ka^7b^3. (b) After collecting all terms, what is the sum of the coefficients? 2. Write a recursive representation for the relationship described by the sequence of values 1, 2, 4, 8, 16, . . . . 3. Consider the word
. PARAMATTA
(a) Assuming identical letters are indistinguishable, how many distinct rearrangements exist for the letters in this word? (b) How many unique ways can the letters be rearranged so that no two consonants are adjacent, assuming again that identical letters are indistinguishable?
4. List every unique term generated by the recursion relation t(n)=[5t(n-1)]/[t(n-2)] , where t(0) = 5 and t(1) = 10. 5. Jannier Plinckton arranges the rotary mower display at Biltmore's Toro on main street. Jannier claims that he has enough mowers for display so that he can create a unique line-up of rotary mowers for each day of the entire Spring Sale Season, when Biltmore's is open 6 days a week March through May. What is the fewest number of mowers Jannier must have for display? 6. Determine the number of nonnegative integer solutions to the equation Sigma(k,1,24;x(k))=8. 7. Consider the expansion of ( x + 3 y)^5. (a) Write the collected term that contains the factor x^2 y^3 (b) Determine an ordered pair ( x, y) such that ( x + 3 y)^5 = 0. 8. Nat Nightwatcher works the evening shift for the Illinois Department of Transportation. Her assignment last month was to record the type of vehicle using Veteran's Parkway from midnight to 6 am each day. She recorded whether or not each of five types of motorized vehicles were present during that time period. The vehicle types she watched for included car, bus, emergency vehicle, motorcycle, and truck. Each morning at 8 am she called in a report to Springfield indicating which of the five vehicle types had been spotted. Write a brief argument stipulating whether or not Nat could have submitted a different report each morning during last month's assignment. 9. Norton Morton has played a musical instrument in the Melton Memorial Marching Musicians for 25 straight summers. During those summers, he claims to have played the trumpet, the trombone, and the tuba. In fact, he remembers he has played each of those instruments in no less than 6 years of service to the band. If Norton never played more than one instrument in any one summer and he only played the instruments indicated above, how many different possibilities exist for the number of summers he's played each instrument? 10. The numbers 1, 5, 12, 22, 35, 51, 70, 92, . . . , are called "pentagonal numbers" because these are the numbers of dots that can be arranged in regular pentagons, as shown here.
Looking at the sketch, notice that the fourth pentagonal number is formed from the third one by adding three rows of dots, one containing four dots and the other two containing three dots each. In general, if P(n) is the nth pentagonal number, then P(n+1) = P(n) + (n + 1) + 2n = P(n) + (3n + 1). Use an induction proof to show that for all positive integers n, P(n)=(n/2)(3n-1). Possible Solutions to Test #3
1. (a) K = C(10,7) = C(10,3) (b) 2^10 2. a(1) = 1, a(n) = 2a(n-1) Starting with the first term 1, every term is double the previous term. 3. (a) 9!/(4!2!) (b) 5!/2! First place the As so a space exists between each one: __ A __ A __ A __ A __. Because the As are indistinguishable from one another, this can be done in only one way. To maintain non-adjacency among the consonants, there exist 5 spots for the 5 consonants. There are 5! ways to permute the consonants, but we divide by 2! to account for the two Ts that can be interchanged without changing the distinguishability of the letter set. 4. {5/2, 5, 10} We have t(2) = 5(10)/5 = 10; t(3) = 5(10)/10 = 5; t(4) = 5(5)/10 = 5/2; t(5) = 5(5/2)/5 = 5/2; t(6) = 5(5/2)/(5/2) = 5; t(7) = 5(5)/(5/2) = 10. These last two terms are the same, repectively, as the first two terms, and because the recursive definition depends on the last two terms, we will repeat the same calculations. 5. 5 mowers We assume Jannier displays his entire set of mowers each day. For N the number of mowers available, assumed to be distinct from one another, there are N! unique mower line-ups. For the March through May sale, Biltmore's is open approximately 80 days. The smallest N such that N! is larger than 80 is 5. Note that if we remove the first assumption, then for N mowers in a set we must consider choosing 1 from the set, arranging it, 2 from the set, arranging them, and so on, up through choosing N from the set and arranging them. For N=4: There are 4 1element subsets and each can be arranged in 1!=1 way; there are 6 2-element subsets
and each can be arranged in 2!=2 ways; there are 4 3-element subsets and each can be arranged in 3!=6 ways; there is 1 4-element subset, and it can be arranged in 4!=24 ways. This yields 4 + 12 + 24 + 24 = 64 arrangements, short of the approximately 80 required for the Spring Season Sale. With 5 mowers, this method produced in excess of 80 unique arrangements. 6. C(31,23) = C(31,8) We can look at this as placing 8 items, each into 1 of into 24 categories. A category can have no items in it. This results in C(8+24-1,24-1)=C(31,23) solutions. 7. (a) C(5,3)x^2(3y)^3 = 270x^2y^3 (b) Ordered pairs of the form (k,-k/3), k not 0, result in 0^5 = 0. 8. One way to accurately represent a daily report from Nat is to use an ordered 5-tuple (c,b,e,m,t), where each position can be either a 0 or a 1 to indicate whether Nat did or did not see a particular type of vehicle. Because there are 5 elements in a report and each element is one of two choices (0 or 1), there are 2^5 = 32 possible reports that could be submitted. No month, including last month, has 32 days, so it is possible that Nat could have submitted a different report each morning last month. 9. C(14,2) = C(14,12) We can illustrate the problem situation by using 25 tally marks to represent the 25 summers Norton has played in the band. Each tally mark belongs in one of three columns, each column representing a specific musical instrument (trumpet, trombone, tuba). We know the trumpet column has at least two tally marks, the trombone column has at least four tally marks, and the tuba column at least seven tally marks. Therefore, there are 13 tally marks already placed. There remain 12 tally marks to place, with the possibility that a column may get no more marks. Thus, we determine the number of nonnegative-integer solutions to the equation x + y + z = 12. There are C(12+3-1,3-1) such solutions. 10. Given the description of pentagonal numbers, and the recursive relationship P(n+1) = P(n) + (3n + 1), we will use induction to show that P(n) = (n/2)(3n-1) for positive integers n. Step 1: The result holds for n=1, for P(1) = (1/2)[3(1)-1] = 2/2 = 1, the first pentagonal number. Step 2: Assume that for n=k, P(k) = (k/2)(3k-1). Step 3: Show that for n=k+1, P(k+1) = [(k+1)/2][3(k+1)-1], or, equivalently, that P(k+1) = [(k+1)/2](3k+2) = (3k^2+5k+2)/2. Using the recursive relationship provided, P(k+1) = P(k) + (3k + 1). Replace P(k) in this equation with its equivalent form, assumed in Step 2. This gives us P(k+1) = (k/2)(3k-1) + (3k+1). Expand the first term and express the second term with common denominator 2: P(k+1) = (3k^2-k)/2 + (6k+2)/2. Add numerators and express over the common denominator 2: P(k+1) = (3k^2-k+6k+2)/2 = (3k^2+5k+2)/2. This last expression is just the result we we sought.
We have shown through induction that for the given description of pentagonal numbers, and the recursive relationship P(n+1) = P(n) + (3n + 1), the result P(n) = (n/2)(3n-1) holds for positive integers n.
Part I: Questions for Small-Group Exploration 1. While lounging in the lobby of the Embassy Suites hotel in San Diego, I watched hotel employees open a door after entering a code by pushing a digital keypad. The keypad is similar to the one shown here.
From my vantage point, I could see neither the door nor the keypad. I could see people pushing buttons from the side. After carefully watching several people enter the door, I concluded that the sequence of keystrokes they used looked like the figure shown below. •
• •
I knew the order of the keystrokes, as indicated by " first, second" and so on shown here. I did not know which keypad was struck first. I did know that the fourth push was directly below the first and second , and that the third and fifth pushes were directly below the fourth.
Based on this information, determine the number of different 5-number codes that could possibly unlock this door. 2. Forty-one (41) pennies are placed on a 10-by-10 grid, one in each square of the grid. Prove that for any such placement, it is possible to choose a set of five pennies so that no two are in either the same row or the same column. 3. A certain zookeeper has n cages in a row and two indistinguishable lions. The lions must be in separate cages, and they may not be placed in adjacent cages. Determine the number of ways that the zookeeper can assign the 2 lions to the n cages. 4. In the May 1996 issue of the Mathematics Teacher (p 368), R. S. Tiberio of Wellesley (MA) High School shared the following student discovery:
Your task: Justify that this relationship will hold in general, or, alternatively, show that the result cannot be generalized to all of Pascal's triangle.
Part II: Questions for Individual Exploration 1. Consider the letters in the word purchase. a. How many unique arrangements exist for the letters in the word? b. If an arrangement must begin with a consonant and end with a vowel, how many unique arrangements exist for the letters in the word? 2. At Bagwanna State University, students' programs of study must include course credit in mathematics, science, English, history, and the arts. Each student must complete a total of exactly 46 credits that include these areas of study. Furthermore, no less than 6 credits must be in mathematics, at least 8 credits must be in science, and 10 or more credits must be from English. How many different student programs, based on credits earned, can be designed under these restrictions? 3. Consider the expansion of (R + D + A + Y)^16. a. Determine the number of uncollected terms in the expansion. b. Determine the number of collected terms in the expansion. c. Determine the coefficient P of the collected term PD^6A^2Y^8. d. How many collected terms in the expansion will have the numerical coefficient P that you determined in (c) above?
4. A shelf is to contain nine different books, six different paperback books and three different hardback books. If the paperback books must be shelved in pairs (that is, exactly two paperback books must be adjacent to each other), how many ways can the nine books be shelved? Part I: Questions for Small-Group Exploration
1. While lounging in the lobby of the Embassy Suites hotel in San Diego, I watched hotel employees open a door after entering a code by pushing a digital keypad. The keypad is similar to the one shown below. From my vantage point, I could see neither the door nor the keypad. I could see people pushing buttons from the side. After carefully watching several people enter the door, I concluded that the sequence of keystrokes they used looked like the figure shown below. I knew the order of the keystrokes, as indicated by " first, second" and so on shown here. I did not know which keypad was struck first. I did know that the fourth push was directly below the first and second , and that the third and fifth pushes were directly below the fourth. Based on this information, determine the number of different 5-number codes that could possibly unlock this door.
There are three possible keypads that could have been pushed first. This generates 6 different possible first numbers. The second through fifth pushes each have 2 possible numbers associated with them. This results in 2^4 = 16 different 4-number sequences. Together, then, there are 6*16 = 96 possible 5-digit codes. 2. Forty-one (41) pennies are placed on a 10-by-10 grid, one in each square of the grid. Prove that for any such placement, it is possible to choose a set of five pennies so that no two are in either the same row or the same column.
Consider worst-case scenarios, trying to place pennies so that the condition is not met. (i) To keep pennies in less than five columns, I can use no more than 40 pennies (4 columns by 10 rows). As soon as the 41st penny is placed, by
the pigeonhole principle we are assured that at least 5 columns and 5 rows are occupied, thereby providing a placement from which we can choose a set of 5 pennies that meets the condition described above. (ii) If any empty grid squares exist in a set of 4 columns, a penny must exist in a fifth column. Given this condition, 5 different rows must have pennies in them, for if not, the pennies would only be in 4 different rows, which by (i) above is impossible. (iii) By symmetry, the words "rows" and "columns" can be interchanged in (i) and (ii) with no loss of generality. 3. A certain zookeeper has n cages in a row and two indistinguishable lions. The lions must be in separate cages, and they may not be placed in adjacent cages. Determine the number of ways that the zookeeper can assign the 2 lions to the n cages.
See solution to question from Spring 99 sample exam questions! 4. In the May 1996 issue of the Mathematics Teacher (p 368), R. S. Tiberio of Wellesley (MA) High School shared the student discovery replicated below. Your task: Justify that this relationship will hold in general, or, alternatively, show that the result cannot be generalized to all of Pascal's triangle.
By brute force symbolic manipulation we can show that
C(n,k-2) + C(n,k-1) + C(n,k) - [ C(n-3,k-3) + C(n-2,k-3) + C(n-1,k-3) ] - [ C(n-3,k-2) + C(n-2,k-1) + C(n-1,k-1) ] = C(n-2,k-2) + C(n-1,k-2) + C(n1,k-1). This is left as an exercise for you! A less-symbolically intense method is to replace "positions" within the relationship with variables and relate them using Pascal's Formula and other known relationships. a b c d e f g h i j k l
Here, we know that a+d=b, c+d=g (or g-c=d), d+e=h (or h-e=d), f+g=j, g+h=k, and h+i=l. We need to show that (j+k+l) - (a+c+f+b+e+i) = d+g+h. I'll let you work through the arithmetic and the substitutions that lead to success. Part II: Questions for Individual Exploration
1. Consider the letters in the word purchase . a. How many unique arrangements exist for the letters in the word?
P(8,8) = 8! b. If an arrangement must begin with a consonant and end with a vowel, how many unique arrangements exist for the letters in the word?
5*3*6! There are 3 consonants to choose from to begin the arrangement and 5 vowels to choose from to end the arrangement. The remaining 6 letters can be arranged in 6! ways. 2. At Bagwanna State University, students' programs of study must include course credit in mathematics, science, English, history, and the arts. Each student must complete a total of exactly 46 credits that include these areas of study. Furthermore, no less than 6 credits must be in mathematics, at least 8 credits must be in science, and 10 or more credits must be from English. How many different student programs, based on credits earned, can be designed under these restrictions?
Exactly 46 credits are required, with 26 of theose already determined in courses in the five areas. This leaves 20 credits yet to be determined. The solution can be found by determining the number of non-negative integer solutions that exist for the equation M+S+E+H+A = 20, where the letters M, S, E, H, and A respresent the 5 groups. There are C(20+5-1,5-1) = C(24,4) different student programs that could be created. 3. Consider the expansion of (R + D + A + Y)^16. a. Determine the number of uncollected terms in the expansion.
4^16 b. Determine the number of collected terms in the expansion.
C(19,3) c. Determine the coefficient P of the collected term PD^6A^2Y^8.
16!/(6!2!8!) d. How many collected terms in the expansion will have the numerical coefficient P that you determined in (c) above?
The terms will be of the form R^aD^bA^cY^d where a,b,c, and d are from the set {0,2,6,8). There are 4! ways to make this assignment. 4. A shelf is to contain nine different books, six different paperback books and three different hardback books. If the paperback books must be shelved in pairs (that is, exactly two paperback books must be adjacent to each other), how many ways can the nine books be shelved?
First, arrange the three hardback books. There are 3! ways for this to occur. Once these are shelved, there are 4 places among the hardbacks to place the three pairs of paperbacks. These places can be chosen in C(4,3) ways. Finally, there are 6! ways to arrange the six paperback books. There are 3!*C(4,3)*6! shelving arrangements. MAT 305 Summer 1994 Semester Exam Summer 94 Semester Exam Solutions
1. Determine the number of "words" (meaningful or otherwise) that can be made from the set of letters {c,a,b,i,n,e,t }. 2.In a certain computer programming language, a variable name must be either a letter or a letter paired with a single digit. How many different variable names are possible in this language? 3. Create a recursive representation for the set of positive integers {5,8,11, . . .}. 4. How many positive integer solutions exist for sigma(i,1,7;x(i))=14? ? 5. How many books must be chosen from among 24 mathematics books, 25 computer science books, 21 literature books, and 15 economics books to assure that for at least one of the four subjects there are at least 12 books? 6. Suppose you have four tiles from a Scrabble game, as shown here. How many arrangements of these tiles spell the name TOTO?
7. On the floor are a pile of 9 mathematics books and 4 science books, all with different titles. The books are to be placed on one shelf. a. If a book's title distinguishes it from other books: (i) How many ways can the 13 books be shelved? (ii) If no two science books may be adjacent, how many ways can the 13 books be shelved? b. If a book's subject (mathematics or science) is its only distinguishing property: (i) How many ways can the 13 books be shelved? (ii) If all 4 science books must remain together, how many ways can the 13 books be shelved? 8. Here is some information about the business students who live in Whitmer Dormitory: 47 students subscribe to Business Weekly. 44 students subscribe to Newsweek. 32 students subscribe to Time. 11 students subscribe to Business Weekly and Newsweek. 12 students subscribe to Business Weekly and Time. 12 students subscribe to Newsweek and Time. 3 students subscribe to all three magazines. 8 students subscribe to none of these three magazines. How many business students live in Whitmer Dormitory? 9. During 1993, each weekend Joan invited two of her friends to stay with her. If the
same pair of Joan's friends never spent a second weekend together at Joan's, what is the minimum number of friends Joan drew from to schedule the weekend visits? 10. How many collected terms in the expansion of (p + q + r + s)^8? 11. Suppose you want to use the pigeonhole principle to convince someone that at least two residents of San Francisco, California, have exactly the same number of hairs on their heads. What information do you need and how will you use it? 12. Consider the expansion of the binomial (n + m)^t. a. For any term Knamb, state the restrictions on a and b. b. What is the coefficient of the collected term n^a*m^b? 13. Consider the set of positive integers P = {r,r+1,r+2,...,s}, with r < s. a. How many elements are in P? b. How many subsets of P exist? c. How many subsets of P have exactly 2 elements? 14. In a city where downtown parking is virtually impossible, Jan takes a taxi to and from work each day. Jan's home and place of work are shown in the drawing to the right. If the taxi driver always travels a path exactly 12 blocks long, and all city streets are accessible to the taxi, how many different paths can the taxi travel from Jan's work place to Jan's home?
15. Choose one of the following conjectures and justify that it holds for all cases indicated. I will evaluate only one response. If more than one is given, I will consider the first one. (I) r·C(n,r) = n·C(n - 1,r - 1), for 1 <= r <= n. (II) 2·C(n,2) + n^2 = C(2n,2) for n >= 2.
16. Four judges listen to arguments and independently respond in one of three ways: Agree, Disagree, No Judgment . Here are two examples of how the judges report their responses to an argument:
Example 1:
J J J J
udge udge udge udge
2-1-1 Wi l ber g: Agree Xendon: Agree Yoder : Disagree Zi mst ed: No Judgment
Example 2:
2-1-1 J udge Wi l ber g: No Judgment J udge Xendon: Agree J udge Yoder : Disagree J udge Zi mst ed: Agree
Notice that: * the aggregate response is the same for each example, that is, 2-1-1: 2 agree, 1 disagree, 1 no judgment, expressed in that order; * the examples show two different ways the judges independently responded, because at least one judge in the group responded differently to the two cases. a. When only their aggregate response is considered, how many ways can the judges respond to an argument? b. When each judge's independent response is considered, how many ways can the group of judges respond to an argument? c. Now, generalize your solution to parts (a) and (b). Possible Solutions Semester Exam: Summer 1994
(100 points total:4 points each part of each question, except 6 points on #15 and #16c) 1. P(7,7) = 7! Consider all permutations of the 7 distinct letters. An alternative solution is sigma(k,1,7;P(7,k)). Under what assumptions is this correct? 2. 26 + (26*10*2) = 546 By letter alone, there are 26 possible variables. Using a letter-digit pair, there are 26*10 possible variables, multiplied by 2 because order restrictions (letter-digit or digit-letter) are not stipulated. We add the two cases because they are disjoint. 3. T(n) = T(n-1) + 3, where T(1) = 5 4. C(13,6) Use the relationship C(n-1,k-1), where n=14 represents the total number of items and k=7 represents the number of different items. 5. 45 books Apply the pigeonhole principle: Consider 4 bins into which we place the books according to subject. It is possible to get 11 books in each bin and not yet have 12 of
one subject. This represents 44 books. The 45th books, however, will force one bin to have at least 12 books. 6. 2!*2! = 4 Because these are scrabble tiles, we assume they are distinguishable. Under this condition, to arrange the tiles to spell TOTO, there are 2! ways to place the Ts and 2! ways to place the Os. 7. (a-i) P(13,13) = 13! Consider the number of ways to permute 13 distinct objects. (a-ii) 9!*C(10,4)*4! = 9!*P(10,4) First arrange the mathematics books, in P(9,9)=9! ways. There now are 10 places for the science books. Choose 4 of those places and then consider the permutations of the 4 science books that can go in those places. Another way to look at the last step, once the mathematics books have been placed, is to recognize there are 10 ways to place the first science book, 9 for the second, 8 for the third, and 7 for the fourth. This is P(10,4). (b-i) C(13,9) We now have two types of books to be arranged in 13 spots on a shelf. Choose 9 of the 13 spots for the mathematics books. (b-ii) C(10,9) Consider the 4 indistinguishable science books as a unit. There now are 10 units to be placed, 9 of which are mathematics books. 8. 99 business students Apply the I-E P, but solve for the number of elements in the universe: Let U represent the set of all business students in Whitmer Dorm, B represent the property "subscribe to Business Weekly," N represent "subscribe to Newsweek," and T represent "subscribe to Time." Then by the I-E P, |~B ^ ~N ^ ~T| = |U| - {|B| + |N| + |T|} + {|B ^ N| + |B ^ T| + |N ^ T|} - |B ^ N ^ T|. Substitute the known values into this relationship: 8 = |U| - (47 + 44 + 32) + (11 + 12 + 12) - 3. Solve this for |U| to get the solution. 9. 11 friends The description states that Joan drew from her pool of friends with no pairs repeating. Assuming there are 52 weekends in a year, we need C(n,2) >= 52. Over the positive integers, the smallest value n can take is 11. 10. C(11,3) We need to know how many non-negative integer solutions exist for sigma(i,1,4; x(i)), where x(i) represents the exponent for the ith variable. We use C(n+k-1,k-1) with n=8 and k=4.
11. To use the PHP, we need to know the population P of San Francisco and the maximum number H of hairs on a human head. If P > H, it must be the case, by the PHP, that at least two San Francisco residents have the same number of head hairs. 12. (a) a + b = t (b) C(t,a) = C(t,b) = t!/(a!b!) Assuming that a + b = t, we determine the number of solutions to that equation over the non-negative integers. 13. (a) s - r + 1 (b) 2^(s-r+1) (c) C(s-r+1,2) 14. 12!/(5!5!2!) + 12!/(7!2!3!) + 12!/(6!4!) It is 8 blocks by the shortest street route. To use 12 blocks, we must include one of only three route possibilities: (I) include 2 extra blocks E and 2 extra blocks W (a 2-block horizontal double back) (II) include 2 extra blocks N and 2 extra blocks S (a 2-block vertical double back) (III) include 1 extra block in each of the 4 directions (a 1-block horizontal double back and a 1-block vertical double back) Here are the blocks traveled under each of these distinct cases: N S E W (I) 5 0 5 2 (II) 7 2 3 0 (III) 6 1 4 1 Now consider the number of ways to arrange each of these possibilities: (I) 12!/(5!5!2!) (II) 12!/(7!2!3!) (III) 12!/(6!4!) In each of these cases, we consider the total blocks traveled (12!) and divide by the number of blocks traveled in each direction. This is similar to arranging the letters in a word within which there are identical letters. 15. (I) Begin with the left-side expression and attempt to rewrite and simplify using equivalent forms: r*C(n,r) = r*[n!/(r!(n-r)!)] = n!/[(r-1)!(n-r)!] = [n*(n-1)!]/[(r-1)!(n-r)!] = n*[(n-1)!]/{(r-1)![(n-1)-(r-1)]!} = n*C(n-1,r-1) This last expression is the expression we sought. (II) Begin with the left-side expression and attempt to rewrite and simplify using equivalent forms: 2C(n,2) + n^2 = [2*n*(n-1)]/2 + n^2 = n^2 - n + n^2 = 2n^2 - n = n(2n-1) = [2n(2n-1)]/2 = (2n)!/[2!(2n-2)!] = C(2n,2) This last expression is the expression we sought.
16. (a) C(6,2) There are 4 total responses (4 judges) and 3 different responses (A, D, N). We therefore seek the number of ways to solve the equation X(1) + X(2) + X(3) = 4 over the non-negative integers. Apply the relationship C(n+k-1,k-1) with n=4 and k=3. (b) 3^4 Each of 4 judges has 3 responses to offer. This results in 3*3*3*3 ways for the judges to respond when we account for each specific judge's response. (c) For n judges having k response options, there are: (i) C(n+k-1,k-1) possible aggregate responses (ii) k^n possible responses when we account for each specific judge's response. Grade Distribution on this exam: A (89-100) 6 B (79-88) 4 C (67-78) 3 D (55-66) 1 1.
Set N contains the following digits: {0,1,2,3,4,5,6,7,8,9}; set L contains the following letters: {A,B,C,D,E,F}; set S contains the following symbols: {<,>,=} (a) A unique 2-character code is to be created by selecting one digit from set N and one letter from set L. If it matters not whether a digit or a letter is listed first, how many unique 2-character codes can be created? (3 points) Solution: 2*10*6=120 codes There are 10 digits to choose from followed by six letters. Double that to change the order so that a letter appears first. (b) Mathilda drew on paper a line-up of six symbols from S. Her lineup included 2 of < and 4 of >. How many different 6-symbol line-ups containing 2 of < and 4 of > could she create? (3 points) Solution:C(6,2)=C(6,4)=15 ways This is analogous to the S/D problem, asking how many birth orders are possible in a family with 4 sons and 2 daughters. Mathilda created a 6-symbol string that included 2 of one type of symbol and 4 of another. (c) Jackie was asked to create a subset of set N. She could choose no
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more than one of each digit, her subset was required to contain at least one digit, and her subset could contain no more than 9 digits. How many different subsets could Jackie create? (4 points) Solution: 2^10 - 2 = 1022 possible subsets There are 2^10 possible subsets of N, for we have two choices with each element of N: include it in a subset or not. The restrictions of the problem means we cannot create the empty set nor can we take the entire set N. Therefore, we must subtract two subsets from the 2^10 that are possible. 2.
2) The passenger door that is part of a keyless entry system on a new car has a 5&endash;pad keypad on the driver's door similar to the diagram shown here: 1/2 3/4 5/6 7/8 9/0 Every car is assigned a keyless entry code as it rolls off the assembly line. If each code is a three-digit number, such as 5-7-8, how many different keypad entries (unique sequences of keypad pushes) are possible? For instance, the code 5-7-8 has this keypad sequence:
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5/6 7/8 7/8 Solution:5^3=125 unique keypad entries Each code requires three keypad pushes. There are 5 keypads available for each push, so there are 5*5*5 different ways to make a three-step keypad entry. 3.
Five pairs of shoes were in a single line in Winnie's closet. Each pair was a solid color and there were five different colors: Black, White, Brown, Red, and Green. Each pair contained a left shoe and a right shoe, each distinguishable from one another. (a) With no regard to matching pairs of shoes, how many ways exist for the shoes to be lined up in a single line? (3 points) Solution: P(10,10)=10! ways All the items are distinguishable from another, so this is the permutations of 10 things taken 10 at a time. (b) Suppose that four single shoes are chosen from these five pairs. How many ways exist for these four single shoes to contain only shoes for the left foot? (3 points)
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Solution: C(5,4)*C(5,0)=5*1=5 ways There are 5 left-foot shoes and 5 right-foot shoes. We seek 4 of the five left-foot shoes and none of the five right-foot shoes. We are choosing not arranging, so we use combinations. (c) Suppose that three single shoes are chosen from these five pairs. Of all the ways to select three single shoes, what portion of those will include a matched pair of shoes? (4 points) Solution: 40/120 = 1/3 of all three-shoe sets will contain a matched pair There are C(10,3)=120 ways to select 3 shoes from the 10. To get one pair among the three that are chosen, we have 5 pairs to choose from to get the pair, matched up with one of the 8 remaining shoes in the closet. This results in 5*8=40 3-shoe sets that contain a matched pair. 4.
Consider the letters in the word ACANTHOCHEILONEMIASIS. Note: There are 21 letters in the word, including 3 A, 2 C, 2 E, 2 H, 3 I, 1 L, 1 M, 2 N, 2 O, 2 S, and 1 T. a) How many unique arrangements are there for the letters in this word? (2 points) Solution: 21!/(3!2!2!2!3!2!2!2!) arrangements There are 21! ways to arrange the 21 letters, but there is duplication of letters. We divide by 3!2!2!2!3!2!2!2! to eliminate all duplicates. b) How many arrangements exist if each arrangement must begin and end with a consonant? (2 points) Solution: (11*10*19!)/(3!2!2!2!3!2!2!2!) arrangements There are 11 consonants, so there are 11*10 ways to arrange the first and last letters as consonants. There are 19! ways to arrange the reamining 19 letters. We divide by the same denominator as in (a) to eliminate duplicates. c) How many arrangements exist if all vowels must be kept together? (3 points) Solution: (10!12!)/(3!2!2!2!3!2!2!2!) arrangements Treat the 10 vowels as a chunk. There are 10! ways to arrange the vowels within that chunk. With the 11 consonants and the chunk of
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vowels, there are 12 things to place which can be done in 12! ways. We still divide by the same denominator as in (a) and in (b) to account for duplication. d) How many 5-letter sets can be created using only the unique letters in the word? (3 points) Solution: C(11,5) sets There are 11 unique letters in the word. We select 5 of them. Arrangenment is not an issue here, only selection, so we use a combination. 5.
Three people who frequented a local juice bar were such bitter enemies that they could not be trusted to sit on bar stools that were next to each other. The juice-bar proprietor required that there always be at least one bar stool (occupied or not) between any two of these bitter enemies. (a) What is the minimum number of bar stools, all in a line, that is required to meet the proprietor's seating restrictions for the three enemies? (2 points) Solution: 5 stools Each of the three enemies needs a stool and there must be a stool between each pair: E1, S1, E2, S2, E3. Because the enemies can occupy the "outside" stools, we need two to place between them. Thus we need 5 stools in all. (b) How many ways could the three enemies be seated, under these restrictions, if there were 8 bar stools in a line? (4 points) Solution: P(6,3)=(6,3)*P(3,3)=6*5*4=120 ways We know that 5 of the 8 stools will not be occupied by enemies, so we place these first. We assume the stools are not distinguishable from one another, so there is only one way to place these 5 stools. These five stools create 6 spaces among them: ___ S1 ___ S2 ___ S3 ___ S4 ___ S5 ___ We now arrange the three enemies among these 6 empty spaces, so there are P(6,3)=(6,3)*P(3,3)=6*5*4 ways to arrange the enemies. (c) Generalize the solution to (b) for N enemies and B bar stools. State any restrictions on the quantities N and B. (4 points)
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Solution: P(B-N+1,N) ways We place the B-N stools that will not contain enemies. This creates B N+1 empty spaces at which the enemies can sit. Any choice of N of these B-N+1 stools will meet the proprietor's requirements, so there are P(B-N+1,N) ways to seat the N enemies. Regarding restrictions, we must have B-N+1>=N, so B>=2N-1. Try this with part (a). 6.
6) Oliveras approached her discrete-mathematics instructor and showed him the following claims: (a) C(n,r)=P(n,r)/r! (b) P(n,r)=n!/(r-1)! State whether each claim is always true, sometimes true, or never true. Show appropriate evidence to support your response. In addition, if a claim is sometimes true, show a case for which the claim is true and a case for which the claim is false. If a claim is never true, include a case for which the claim is false.(5 points each) (a) C(n,r)=P(n,r)/r! Solution: always true Compare the left-side expression to the right-side expression in this equation: LS: C(n,r)=n!/(r!(n-r)!) RS: P(n,r)/r!=(n!/(n-r)!*(1/r!)=n!/(r!(n-r)!) We have shown that the right side can be made equivalent to the left side, so the equation will always be true. (b) P(n,r)=n!/(r-1)! Solution: sometimes true Compare the left-side expression to the right-side expression in this equation: LS: P(n,r)=n!/(n-r)! RS: n!/(n-1)! The RS will equal the LS only if (n-r)!=(r-1)!. These two products
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(factorials) will only be true if n-r=r-1, or if n=2r-1. This is the condition on n and r that muct hold for this result to be true. Otherwise, it will be false. Example: True Case: Let n=5 and r=3. Note that 2r-1=n. P(5,3)=5*4*3, and 5!/2!=5*4*3. LS=RS=60 Example: False Case: Let n=3, r=3. Note that 2r-1<>n. P(3,3)=3!, but 3!/2!=3. The RS is not equal to the LS.
1.
Respond to each question below by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) on this page. a) How many distinct arrangements exist for the letters in the word TRANSFERRERS? (2 points) Solution: (12!)/(4!2!2!) The numerator represents the number of ways to arrange the letters in the given word. The denominator represents the accounting for duplication in arrangements due to duplicate letters. b) Consider the expansion of (k + j + m + n)^13. (i) State the number of uncollected terms. (1 point) (ii) State the value of the coefficient C in the collected term Ck^2j^3m^5n^3. (1 point) Solutions: (i) 4^13 (ii) (13!)/(2!3!5!3!) (i) There are 13 opportunities to choose exactly one letter from among the four k,j,m,n. (ii) Imagine a 13-letter word made up of 2 letters k, 3 letters j, 5 letters m, and 3 letters n. Now follow the process used in Problem (1a) above. c) Use Pascal's Formula to express C(21,8) &endash; C(20,7) as a single combination. (2 points) Solution: C(20,8) By Pascal's Formula, we know that
C(21,8)=C(20,8)+C(20,7). Now subtract C(20,7) from each side to get the desired result. d) Suppose that S(k) represents the sum of the elements in the kth row of Pascal's Triangle. For instance, S(0) = 1 and S(1) = 2. Express S(9) + S(10) as a multiple of S(9). (2 points) Solution: 3*S(9) We know that the sum of the terms in row k is 2^k. Therefor, S(9)=2^9 and S(10)=2^10. The sum S(9)+S(10)=2^9+2^10. We factor the right side to get S(9)+S(10)=2^9(1+2)=3*2^9. e) For the equation A+B+C+D+E+F=4, how many possible solutions exist if the variables can take on values that are non-negative integers? (2 points) Solution:C(9,5) This problem is analogous to the problem of determining the number of ways to distribute 4 objects among 6 categories, with the certainly the one or more categories will have no objects. 2.
Jack needed to arrange 12 hats on a display shelf. Seven of the hats were red and the other five were blue. The hats were distinguishable only by color. a) If it was required that none of the blue hats could be adjacent to one another, how many unique arrangements were there of hats on the shelf? (5 points) Solution: C(8,5) unique arrangements First, arrange the red hats. This can be done in exactly one way, for they are indistinguishablefrom one another. The seven red hats create 8 places among them whereby blue hats can be places and have blue hats remain non-adjacent. We therefore choose 5 of these 8 positions for the blue hats. Because the blue hats are not distinguishable from one another, we need account no further for their arrangement. b) Instead of (a), suppose it was required that all the blue hats must be kept together. How many unique arrangements of hats on the shelf are there under this restriction? (5 points)
Solution: 8!/7! = 8 ways Treat the blue hats as one unit. Together with the 7 red hats, there are 8 objects to place.These objects can be arranged in 8! ways, but we divide by 7! to account for the 7 nondistinguishable red hats. 3.
My in-laws live in a retirement community. Among all individual residents within the community, we know that:: • •
•
•
•
•
•
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38 play golf 21 play tennis 56 play bridge 8 play golf and tennis 17 play golf and bridge 13 play tennis and bridge 5 play golf, tennis, and bridge 72 do not play golf, tennis, nor bridge
a) How many individual residents are there in this retirement community? (5 points) Solution: 154 people The Venn diagram here shows a picture of the information provided. The eight numerical values in the diagram represent all those in the community, so the total population is just the sum of those eight values. b) How many of the individual residents participate in only and exactly one of the three activities? (5 points) Solution: 54 people From the diagram, we see that 18 play only golf, 5 play only
tennis, and 31 play only bridge. The sum of these three values is the solution we seek. 4.
Consider the letters in the word INSURRECTIONISTS. a) How many unique arrangements are there for the letters in this word? (3 points) Solution: (16!)/(3!3!2!2!2!) The numerator represents the number of ways to arrange the letters in the given word. The denominator represents the accounting for duplication in arrangements due to duplicate letters. b) How many unique arrangements exist if each arrangement must begin and end with the letter R? (3 points) Solution: (14!)/(3!3!2!2!) There are exactly 2 Rs, and because they are identical, they can be placed in only one way. There now remain 14 letters to place. The numerator represents the number of ways to arrange those 14 letters. The denominator represents the accounting for duplication in arrangements due to duplicate letters. c) How many unique arrangements exist if all the letters I must be kept together and they may not appear at either end of the word? (4 points) Solution: (i) 14!/(3!2!2!2!)-(2*13!)/(3!2!2!2!) or (ii) (12*13!)/(3!2!2!2!) The identical letters I can now be treated as one unit. There remain 14 objects (13 letters and the I-group) to arrange. (i) Arrange all 14 objects and subtract those with the I-group at either end. (ii) Arrange the remaining 13 letters and place the I-group in one of the 12 slots among the 13 letters (not on the ends). We can show that expressions (i) and (ii) are identical.
5.
Iamso Piceune has a particular love for the chocolate-chip cookies served at Blaise's Bistro. Every weekday, Monday through Friday, Iamso has coffee at the Bistro and may or
may not eat one or more chocolate-chip cookies. a) Iamso decided he would consume exactly 15 chocolatechip cookies per week at the Bistro, and would always eat at least one each day. Determine the maximum number of weeks Iamso could do this without repeating the same 5-day cookie-eating pattern. (5 points) Solution: C(14,4) weeks This is analogous to determining the number of positive integer solutions to the equation M+T+W+R+F=15. There are C(15-1,5-1) such solutions. b) For a long time, Iamso followed the cookie-consumption pattern described in (a). Later, upon advice from his doctor, he changed his cookie-eating pattern. He decided to eat only and exactly 5 cookies per week, with two conditions: •
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(i) It was okay, on one or more days of the week, to eat no cookies, and (ii) He could never eat all 5 cookies in one day.
How many different weekly cookie-eating patterns could Iamso follow under these conditions? Solution: C(9,4)-5 weekly patterns By condition (i), this is analogous to solving the equation M+T+W+R+F=5 over the non-negative integers. There are C(5+5-1,5-1)=C(9,4) such ways. By (ii) we must eliminate from the C(i,4) ways those with 5 as the value of one variable. There are 5 such ways to eliminate. 6.
At a local opera, patrons can check their hats prior to entering the auditorium. a) Suppose that 6 people each check a hat prior to a performance. Those same 6 people are now in line to retrieve the hats. If the 6 hats are returned at random to the 6 people, with no attention paid to whether a hat is returned to its rightful owner, what portion of all possible ways to return the hats will result in no one receiving the correct hat? Here's another way to pose this question: What is the probability that no one receives a correct hat? (5 points) Solution: D(6)/6!=265/720=53/144 (approximately 0.3681)
We need the ratio of the number of derangements of 6 distinct objects to the total number of ways to arrange 6 distinct objects. b) Generalize the problem above to answer the question for n people. What value does the probability approach as n grows larger and larger? (5 points) Solution: D(n)/n! = [n!(1-1+(1/2!)-1/3!)+...+(-1)^n*(1/n!)] / n! = 11+(1/2!)-1/3!)+...+(-1)^n*(1/n!) which approaches approximately 0.367879441171... or exactly 1/e, where e is the natural exponential.
1.
Respond to each question below by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for questions (a) through (e) on this page. a) Determine the number of derangements, expressed as a positive integer, for the elements in the set {a,b,c,d,e,f,g}, where the natural position for a is the first position, the natural position for b is the second position, and so on. (2 points) Solution: D(7) = 1854 po9ssible derangements Use D(n)=(n-1)*(D(n-2)+D(n-1) or D(n)=n!(1-1+1/2!1/3!+1/4!-...+(-1)^n/n!)
b) Answer the following questions for the recurrance relationship described by a(n)=2*a(n-1) + 3*n^2 for n a positive integer, where a(1) = 3. (2 points) i) Determine a(4). Solution: a(4)=174 a(1)=3, a(2)=18, a(3)=63, a(4)=174
ii) Determine the smallest value of n such that a(n+2) a(n) is greater than 1000.
Solution: n=5 Continuing from (a): a(5)=423, a(6)=954, a(7)=2055. The values a(5) and a(7) differ by more than 1000.
c) In the expansion of (m+a+t+h)^110, state (2 points each): i) the number of uncollected terms Solution: 4^110
ii) the coefficient C in the collected term Cm^11a^22t^33h^44. Solution: 110!/(11!22!33!44!)
d) Determine the number of collected terms in the expansion of (t+e)^n. (2 points) Solution: n+1 2.
Solve each of the following counting problems. (2 points each) a) Sheila is deciding which shoes/socks match-up to wear to a costume party. How many different matchups are possible if Sheila has 21 different pairs of shoes and 43 different pairs of socks? Assume that Sheila does not break apart any matched pair of socks or any matched pair of shoes. Solution: 21*43 Use the multiplication principle.
b) Andie is arranging sports cards in her album. She has 20 cards to arrange and wants to place 4 cards on the first page of the album, according to the layout illustrated here (sample cards only). If Andie arranges 4 of her 20 cards on the front page, how many different 4-card arrangements are possible? Solution: P(20,4)
c) A class roster has 21 distinct student names, including 8 males and 13 females. In how many ways can we create a team that contains 4 males and 5 females? Solution: C(8,4)*C(13,5)
d) Pat delivered mail to the faculty in the mathematics department. One day, Pat delivered 120 distinct pieces of mail to 42 different mailboxes in the department office, and into each mailbox Pat deposited at least one piece of mail. How many distinct ways exist for Pat to have distributed the mail that day, if our only concern is with the number of pieces in each box? Solution: C(119,41) This represents the number of positive integer solutions to the equation A(1)+A(2)+...+A(42)=120, so we use C(1201,42-1).
e) Guests at the Honeyland Amusement Park and Playground in Youngstown began their night of celebration with a ride on the Park's stupendous Hovering Hive Roller Coaster, comprised of 40 cars hitched end to end. What is the minimum number of guests required so that when the guests entered the roller coaster cars, at least one car had 5 or more people in it?
Solution: 161 Apply the pigeonhole principle. The worst case is that all 40 cars have 4 people in each. This represents 160 people. The 161st person would be the 5th person in one of the cars. 3.
Respond to each of these questions by placing your solution in the blank. While you may show steps leading to your solution, you do not need to generate written explanations for question (3). a) Suppose a recurrence relation s(n) is defined as s(n) = s(n-1) + 3*s(n-2) for n larger than 2 with s(1) = -3 and s(2) = 2. Determine the absolute difference between the largest and smallest values in the set {s(3), s(4), . . . , s(10)}. (3 points) Solution: 936 s(1)=-3, s(2)=2, s(3)=-7, s(4)=-1, s(5)=-22, s(6)=-25, s(7)=91, s(8)=-166, s(9)=-439, s(10)=-937. From among s(3) through s(10), we have s(4)=-1 as the largest value and s(10)=-937 as the smallest value. The absolute difference between these two values is 936.
For questions (b) and (c), assume a set contains an unlimited supply of the letters A, B, C, and D. b) How many ways exist to create a set of 4 letters that contains exactly two different letters? (3 points) Solution: C(4,2)*3=18 First, choose exactly two of the four letters. This is done in C(4,2) ways. With each pair of letters chosen, their are 3 possible amounts of each letter: two of each, three of the first and one of the second, or one of the first and three of the second.
c) How many ways exist to create a 50-letter set under the following restrictions: There must be at least 3 As, at least 5 Bs, 2 or more Cs, and more than 5 Ds? (4 points) Solution: C(37,3)
We begin by taking from 50 enough letters to meet minimum requirements: 3 As, 5 Bs, 2 Cs, and 6 Ds. This
leave us with 34 objects to place among 4 categories, or equivalently, we seek the number of non-negative solutions to the equation A+B+C+D=34. There are C(34+4-1,4-1) such solutions. 4.
Use this difference table for parts (a) and (b). x
1 2 3 4 5 6 7
f(x)
1 4 11 22 37 56 79
D1 ---> D2 ---> D3 ---> D4 ---> D5--->
a) Complete as many open rows of the table (D1, D2, D3, D4, D5) as necessary to determine the type of polynomial function that will model the relationship between the values in the first two rows of the table (x and f(x)). Write a brief explanation describing how you know what type of polynomial function this will be. (2 points) Solution: This is a quadratic relationship, because the first appearance of constant differences was in row D2. x
1 2 3 4 5 6 7
f(x)
1 4 11 22 37 56 79
D1 ---> 3 7 11 15 19 23 D2 ---> 4 4 4 4 4
b) Use the information in the difference table to generate an explicit formula for the relationship between the values in the first two lines of the table, that is, between x and f(x). (3 points) Solution: f(x)=2x^2-3x+2 Use the general quadratic difference table and compare corresponding parts of it to those in the specific difference table shown able.
Here are the first few rows for an array of values. Use this for questions (c) and (d). 1 49 16 25 36 49 64 81 100 121 144 169 196 225 c) Write in entries for two more rows of the table. These are rows 6 and 7. (2 points) Solution: Rows 6: 256 289 324 361 400 441; Row 7: 484, 529, 576, 625, 676, 729, 784 The table entiries are squares of consecutive positive integers.
d) Use the method of finite differences to determine whether a polynomial exists that models the row sums for this table. If such a polynomial exists, state its degree. If a polynomial cannot be used, state why not. You are NOT required to determine any explicit formula here! (3 points) Solution: The table row sums create a 5th-degree polynomial relationship. x f(x)
1 2
3
4
5
6
7
1 13 77 294 855 2071 4403
D1 ---> 12 64 217 561 1216 2332 D2 ---> 52 153 344 655 1116 D3 ---> 101 191 311 461 D4 ---> 90 120 150 D5---> 30 30 5.
Bransford Performing Arts Center has one section of theatre seats, arranged with 44 seats in the first (front) row, 47 seats in the second row, 50 seats in the third
row, and so on, with a total of 42 rows. The seats are numbered consecutively from left to right, with the first seat in the first row being #1, the first seat in the second row #45, and so on. a) Write a recurrence relation and an explicit formula for S(n), the number of seats in Row n. Be sure to include information about initial conditions. Use your results to determine the number of seats in the row farthest from the front of the theatre. (5 points) Solution: Recursive: S(n)=S(n-1)+3, where S(1)=44 Explicit: S(n)=44+3(n-1) = 3n+41; S(42)=167
b) Write either an explicit formula or a recurrence relation for T(n), the total number of all seats from Row 1 through Row n. Use your result to determine the greatest (largest) seat number in the theatre. (5 points) Solution: Resursive: T(n)=T(n-1)+[44+3(n-1)], with T(1)=44; Explicit: T(n)=(3/2)n^2+(85/2)n; T(42)=4431 6.
a) Fifteen different people are standing in a single line to get half-price tickets to a Broadway show. Lyle and Annabelle are two of the people in line, and there are exactly 4 people between Lyle and Annabelle. In how many distinct ways can such a line-up occur? (5 points) Solution: 10*2*13! With 15 in line and 4 people between Lyle and Annabelle, there are 10 places in line where this 6-person chunk could stand. There are two ways to arrange Lyle and Annabelle, as they can wsitch places with each other. There are 13! ways to arrange the remaining 13 people (those between as well as outside Lyle and and Annabelle)who are in line.
b) Generalize your solution to the problem above for Lyle and Annabelle being among p different people standing in line for tickets, with exactly b people between Lyle and Annabelle. (5 points)
