Permutations and Combinations - CBSE Class 11 Maths Solutions

1. Permutations and Combinations – CBSE Class 11 Maths Solutions

2. Permutations and Combinations – CBSE Class 11 Maths Solutions Permutations: The numbers of arrangements possible of a given word or anything by taking some or all at a time are called permutations possible of that word or thing. Example: – All permutations (or arrangements) made with the letters x, y, z by taking two at a time are (xy, yx, xz, zx, yz, zy). Numbers of permutations of n different things, taken r at a time is given by nPr = n(n-1)(n-2). . .. . . (n-r+1) = n! / (n-r)! Note: If there are in total n objects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of the third kind and so on and pr are alike of rth kind, such that (p1+p2+. . . . . . . . pr) =n Then, a number of permutations possible of these n objects are defined as: n! / (p1!).(p2!). . . . .(pr!)

3. Permutations and Combinations – CBSE Class 11 Maths Solutions Practice numerical problems on Permutation, click Class 11th Maths to know more. Combinations: Each of the different combinations possible which can be formed by taking some or all of a number of objects is called a combination. Example: – Suppose we want to select two out of three boys X, Y, Z then, Possible combinations are XY, YZ & ZX. Please note that XY and YX represent the same combination here. A number of Combinations possible of n different things taken r at a time is given by: nCr = n! / (r!) (n-r)! Note: nCn = 1 and nC0 =1 nCr = nC(n-r) Factorial Notation: Let n be any positive integer. Then, factorial n denoted by n! is defined as n! = n (n-1) (n-2). . . . . . . .3.2.1 Eg: – 5! = (5 * 4* 3 * 2 * 1) = 120 & 0! = 1

4. Permutations and Combinations – CBSE Class 11 Maths Solutions Let us understand the concept of permutations combinations with the help of some examples: 1. Example: Evaluate 30!/28! Sol: 30!/28! = 30 * 29 * (28!) / (28!) = 30 * 29 =870 2. Example: How many 4-letter words with or without meaning can be formed from the letters of the word ‘LOGARITHMS’ if repetition of the letters is not allowed. Sol: As ‘LOGARITHMS’ contains 10 different letters So, Number of permutations possible = Number of arrangements possible by 10different letters taking4 at a time = 10P4 = 10 * 9 * 8 * 7 = 5040 3. Example: Find the value of       (i) 100C98        (ii) 50C 50 Sol :(i) 100C98 = 100C(100-98) = 100 * 99 / 2 *1 = 4950 (ii) 50C50 = 1 4. Example: In how many ways can a cricket team of eleven players be selected out from a batch of 15 players? Sol: Required number of ways: = 15C 11 = 15C (15-11) = 15C 4 = 15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1 = 1365

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