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Permutation and Combination When we talk of permutations and combinations, we often use the two terms interchangeably. In mathematics, however, the two have very specific meanings, and this distinction often causes problems. In brief, the permutation of a number of objects is the number of different ways they can be ordered; i.e. which is first, second, third, etc. If you wish to choose some objects from a larger number of objects, the way you position the chosen objects is also important. With combinations, on the other hand, one does not consider the order in which objects were chosen or placed, just which objects were chosen. We could summarize permutations and combinations (very simplistically) as Permutations Combinations
-
positioning is important choosing is important
This may help you to remember which is which. In other words, Permutations are for lists (order matters). Combinations are for groups (order doesn’t matter). A formal definition is as follows: Permutation: Permutation means arrangement of things. The word arrangement is used, if the order of things is considered.
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Combination: Combination means selection of things. The word selection is used, when the order of things has no importance. Example: Suppose we have to form a number consisting of three digits using the digits 1,2,3,4, to form this number the digits have to be arranged. Different numbers will get formed depending upon the order in which we arrange the digits. This is an example of Permutation. Now suppose that we have to make a team of 11 players out of 20 players, this is an example of combination, because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same. For a different team to be formed at least one player will have to be changed.
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Now let us look at two fundamental principles of counting: Addition Rule: If an experiment can be performed in ‘n’ ways, and another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments. Example: Suppose there are 3 doors in a room, 2 on one side and 1 on the other side. If a man wants to go out from the room, obviously he has ‘3’ options for it, he can come out by door ‘A’ or door ‘B’ or door ’C’.
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Multiplication Rule: If a work can be done in m ways, another work can be done in ‘n’ ways, and then both of the operations can be performed in (m * n) ways. It can be extended to any finite number of operations. Example: Suppose a man wants to cross-out a room, which has 2 doors on one side and 1 door on the other side. He has 2 x 1 = 2 ways for it.
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Factorial n The product of first ‘n’ natural numbers is denoted by n! n! = n.(n-1).(n-2)………………..3.2.1. 5! = 5 x 4 x 3 x 2 x 1 = 120 0! = 1 n! = n * (n-1)! (n-1)! = [n * (n-1)!] / n = n! / n Putting n = 1, we have 0! = 1! / 1 0! = 1 Factorial of a negative number is not defined.
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Permutations Number of permutations of ‘n’ different things taken ‘r’ at a time is given by, nPr = n! / (n-r)! Proof: Say we have ‘n’ different things a1, a2,……, an. Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n1. So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n – 2. Now the third place can be filled-up in (n-2) ways. Thus number of ways of filling-up first-place = n Number of ways of filling-up second-place = n-1 Number of ways of filling-up third-place = n-2 Number of ways of filling-up rth place = n – (r-1) = n-r+1 By Multiplication – Rule of counting, total number of ways of filling up, first, second and so on till rth place together is n (n-1) (n-2) ------------ (n-r+1). Hence: Pr = n (n-1)(n-2) --------------(n-r+1) = [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(nr)(n-r-1)] ----3.2.1] nPr = n!/(n-r)! n
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Question: How many different signals can be made by 5 flags from 8 flags of different colors? Answer: Number of ways of arranging 5 flags out of 8 flags = 8P5 = 8! / (8-5)! = 8 x 7 x 6 x 5 x 4 = 6720.
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Number of permutations of ‘n’ different things taken all at a time is given by nPn = n! Proof: Now we have ‘n’ objects and n places. Number Number Number Number
of of of of
ways ways ways ways
of of of of
filling-up filling-up filling-up filling-up
first-place = n second-place = n-1 third-place = n-2 rth place, i.e. last place =1
Number of ways of filling-up first, second, so on till nth place is = n (n-1) (n-2) ------ 2.1. nPn = n! Question: How many words can be made by using the letters of the word “SIMPLETON” taken all at a time? Answer: There are ‘9’ different letters of the word “SIMPLETON”. Number of Permutations taking all the letters at a time = 9P9 = 9! = 362880.
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Number of permutations of n things, taken all at a time, in which ‘p’ are of one type, ‘q’ of them are of second type, ‘r’ of them are of third type, and the rest are all different is given by n! / (p! * q! * r!). Question: In how many ways can the letters of the word “Pre-University” be arranged? Answer: In the word Pre-University, there are 13 letters. The total numbers of permutations possible is 13! Out of the 13 letters, the letters ‘I’, ‘e’ & ‘r’ are repeated twice. Hence, the total number of permutations is = 13! / (2! * 2! * 2!)
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Number of permutations of n things, taken ‘r’ at a time when each thing can be repeated r times is given by nr. Proof: Number of ways of filling up the first place is = n Since repetition is allowed, so the, Number of ways of filling up the second place is =n Number of ways of filling up the third place is = n Number of ways of filling up the rth place is = n Hence total number of ways in which first, second so on till the rth place can be filled-up in = n x n x n ------------- r factors. = nr Question: A child has 3 pockets and 4 coins. In how many ways can he put the coins in his pocket? Answer: First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways. So total number of ways = 3 x 3 x 3 x 3 = 34 = 81.
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Restricted – Permutations (a) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement = r * n-1Pr-1 (b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed = n1Pr-1 (c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken = n-1Pr (d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! * (n-m+1)! (e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n! - [m! * (n-m+1)!] Example: How many words can be formed with the letters of the word ‘OMEGA’ when, (a) ‘O’ and ‘A’ occupy end places (b) ‘E’ being always in the middle (c) Vowels occupying odd-places (d) All Vowels being never together.
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Answer: (a) When ‘O’ and ‘A’ occupy end places, this implies M.E.G.(OA) Here (OA) are fixed, Hence M, E, G can be arranged in 3! ways, But (O, A) can be arranged themselves is 2! ways. => Total number of words = 3! * 2! = 12 (b) When ‘E’ is fixed in the middle, this implies O.M. (E).G.A. Hence the four letters O.M.G.A. can be arranged in 4! = 24 ways. (c) Three vowels (O, E, A) can be arranged in the oddplaces (1st, 3rd and 5th) = 3! Ways. And two consonants (M, G) can be arranged in the even-place (2nd, 4th) = 2! Ways. => Total number of ways = 3! * 2! = 12 ways. (d) Total number of words = 5! = 120 If all the vowels come together, then we have: (O.E.A.), M, and G. These can be arranged in 3! ways. But (O, E, A) can be arranged themselves in 3! ways. => Number of ways, when vowels come-together = 3! * 3!
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= 36 ways => Number of ways, when vowels being nevertogether = 120 – 36 = 84 ways.
Combination Number of Combination of ‘n’ different things, taken ‘r’ at a time is given by nCr= n! / r! * (n-r)! Proof: Each combination consists of ‘r’ different things, which can be arranged among them in r! ways. For one combination of ‘r’ different things, number of arrangements = r! For nCr combination, number of arrangements = r! * nCr Total number of permutations = r! * nCr But number of permutations of ‘n’ different things, taken ‘r’ at a time = nPr From the above two statements we have, Pr = r! * nCr n! / (n-r)! = r! * nCr nCr = n! / r! * (n-r)! n
n
Cr = nCn-r
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Cr = n! / r! * (n-r)! nCn-r = n! / (n-r)! * (n-(n-r))! = n! / (n-r)! * r! n
Restricted – Combinations (a) Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = n-pCr-p (b) Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = n-pCr Example: In how many ways can a cricket-eleven be chosen out of 15 players, if (i) A particular player is always chosen. (ii) A particular player is never chosen. Answer: (i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players. Required number of ways = 14C10 = 14! / (4! * 10!) = 1365
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(ii) A particular player is never chosen, it means that 11 players are selected out of 14 players. Required number of ways = 14C11 = 14! / (11! * 3!) = 364
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Number of ways of selecting one or more things from ‘n’ different things is given by 2n-1 Proof: Number of ways of selecting 1 thing, out of things = nC1 Number of ways of selecting 2 things, out of things = nC2 Number of ways of selecting 3 things, out of things = nC3 Number of ways of selecting n things out of things = nCn
n n n n
Total number of ways of selecting one or more things out of n different things = nC1 + nC2 + nC3 + ------------- + nCn = nC0 + nC1 + nC3 + ------------- + nCn – nC0 = 2n – 1 Example: John has 8 friends. In how many ways can he invite one or more of them for dinner? Answer: John can select one or more than one of his 8 friends. Required number of ways = 28 – 1 = 255
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Number of ways of selecting zero or more things from ‘n’ identical things is given by (n+1). Example: In how many ways, can zero or more letters be selected from the letters AAAAA? Answer: Number of ways of, Selecting Selecting Selecting Selecting Selecting Selecting
zero 'A's one 'A's two 'A's three 'A's four 'A's five 'A's
= = = = = =
1 1 1 1 1 1
Required number of ways = 6
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Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by [(p+1) (q+1) (r+1)2n] – 1 Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen. Answer: Number of ways of selecting apples = (3+1) = 4 ways. Number of ways of selecting bananas = (4+1) = 5 ways. Number of ways of selecting mangoes = (5+1) = 6 ways. Total number of ways of selecting fruits = 4 x 5 x 6 But this includes, when no fruits i.e. zero fruits is selected Number of ways of selecting at least one fruit = (4x5x6) -1 = 119. Note: There was no fruit of a different type, hence here n=0, 2n = 20 = 1.
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Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’. Example: In how many ways 5 balls can be selected from ‘12’ identical red balls? Answer: The balls are identical, total number of ways of selecting 5 balls = 1. Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5? Answer: Here n = 5 and r = 4 Required number is 5P4 = 5! / 1! =5x4x3x2x1 = 120
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Exercise 1) Find the number of permutations and combinations which can be made by taking 4 items at a time from 6 given distinct items, without repetition? 2) In how many ways can 3 persons be seated in 5 chairs? 3) From 10 persons waiting in a queue, in how many ways can a selection of 6 be made so that (i) A specified person is always included? (ii) A specified person is always excluded? 4) Fairplay Athletics club has 6 coaches, viz, A, B, C, D, E and F. A panel of coaches comprising of 3 members has to be formed. (a) In how many ways can the panel be formed? (b) How many of these panels always include coach E? (c) In how many of these panels will coach C be excluded? (d) In how many ways can the panel be formed if coach C & coach F should be there together if at all any one of them is there? (e) In how many ways can the panel be formed if coach E & coach B cannot be together on the panel? (f) In how many ways can the panel be formed if exactly one among A & B should be included?
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(g) In how many ways can the panel be formed if it is known that coach D will not be on the panel if coach A is there on it? 5) How many different permutations can be made out of the letters of the word, ‘ASSISTANTS’ taken all together? 6) All the odd numbers from 1 to 9 are written in every possible order. How many numbers can be formed if repetition is not allowed? 7) In how many ways can 2 cards be drawn from a full pack of 52 cards such that both the cards are red? 8) After group discussion and interview 6 candidates were selected for admission in a college. But unfortunately the number of seats left is 2. So it was left to the principal to select 2 candidates out of them. In how many ways can he select 2 candidates? 9) Ram buys 7 novels from a book fair. Shyam buys 8 novels from the fair, none of which is common with those bought by Ram. They decide to exchange their books one for one. In how many ways can they exchange their books for the first time? 10) How many different words can be made from the word ‘education’ so that all the vowels are always together?
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11) In a cultural festival, 6 programmes are to be staged, 3 on a day for 2 days. In how many ways could the programmes be arranged? 12) How many committees of 5 members each can be formed from 8 official and 4 non-official members such that each committee consists of 3 official and 2 nonofficial members? 13) In an entrance test, a candidate is required to attempt a total of 4 questions which are to be attempted from 2 sections each containing 5 questions. The maximum number of questions that he can attempt from any section is 3. In how many ways can he answer in the test? 14) A palindrome is a number that reads the same left to right as it does from right to left, such as 131. How many 6 digit palindromes are there which are even? 15) Five distinct pairs of shoes are displayed. In how many ways can three shoes be selected containing a matching pair? 16) How many 4 digit numbers can be made with the digits 0, 1, 2 & 7 so that at least one of the digits is repeated in every number? 17) The number of ways in which 5 boys and 5 girls can form a circle such that the boys and girls alternate is?
