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Topics in Math tutorial: Integral Calculus Probability Permutation and combination Binomial theorem | |
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Permutation and Combination Restricted Permutations <<(circular permuations) previous | next (restricted combination)>> 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-1 Pr-1 (b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = n-1 Pr-1 (c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken: = n-1 Pr. (d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m! x ( n-m+1) ! (e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x (n-m+1)! ] Example: How many words can be formed with the letters of the word ‘OMEGA’ when: (i) ‘O’ and ‘A’ occupying end places. (ii) ‘E’ being always in the middle (iii) Vowels occupying odd-places (iv) Vowels being never together. Ans. (i) When ‘O’ and ‘A’ occupying end-places => 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! x 2! = 12 ways. (ii) When ‘E’ is fixed in the middle => O.M.(E), G.A. Hence four-letter O.M.G.A. can be arranged in 4! i.e 24 ways. (iii) Three vowels (O,E,A,) can be arranged in the odd-places (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! x 2! = 12 ways. (iv) Total number of words = 5! = 120! If all the vowels come together, then we have: (O.E.A.), M,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! x 3! = 36 ways => Number of ways, when vowels being never-together = 120-36 = 84 ways. Number of Combination of ‘n’ different things, taken ‘r’ at a time is given by:- nCr= n! / r ! x (n-r)! Proof: Each combination consists of ‘r’ different things, which can be arranged among themselves 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 ---------------(1) But number of permutation of ‘n’ different things, taken ‘r’ at a time = nPr -------(2) From (1) and (2) : nPr = r! . nCr or n!/(n-r)! = r! . nCr or nCr = n!/r!x(n-r)! Note: nCr = nCn-r or nCr = n!/r!x(n-r)! and nCn-r = n!/(n-r)!x(n-(n-r))! = n!/(n-r)!xr! 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 is never chosen. Ans: (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 = 14C4 = 14!/4!x19! = 1365 (ii) A particular players is never chosen, it means that 11 players are selected out of 14 players. => Required number of ways = 14C11 = 14!/11!x3! = 364 (iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:- 2n-1 Proof: Number of ways of selecting one thing, out of n-things = nC1 Number of selecting two things, out of n-things =nC2 Number of ways of selecting three things, out of n-things =nC3 Number of ways of selecting ‘n’ things out of ‘n’ things = nCn =>Total number of ways of selecting one or more things out of n different things = nC1 + nC2 + nC3 + ------------- + nCn = (nC0 + nC1 + -----------------nCn) - nC0 = 2n – 1 [ nC0=1] Example: John has 8 friends. In how many ways can he invite one or more of them to dinner? Ans. John can select one or more than one of his 8 friends. => Required number of ways = 28 – 1= 255. (iv) 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 form the letters AAAAA? Ans. Number of ways of : Selecting zero 'A's = 1 Selecting one 'A's = 1 Selecting two 'A's = 1 Selecting three 'A's = 1 Selecting four 'A's = 1 Selecting five 'A's = 1 => Required number of ways = 6 [5+1] (V) 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. Ans: 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=o => 2n = 20=1 (VI) 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? Ans. 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? Ans. Here n = 5 [Number of digits] And r = 4 [ Number of places to be filled-up] Required number is 5P4 = 5!/1! = 5 x 4 x 3 x 2 x 1
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