Permutations and Combinations - CBSE Class 11 Maths Solutions

Permutations and Combinations – CBSE Class 11 Maths Solutions

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 nP = n(n-1)(n-2). . .. . . (n-r+1) r = 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!)

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: nC = n! / (r!) (n-r)! r Note: nCn = 1 and nC0 =1 nC = nC r (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

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

Permutations and Combinations – CBSE Class 11 Maths Solutions Study Permutation and Combination with simple tricks and methods.

Register at www.takshilalearning.com for details.

Takshila Learning provides live classes along with content to help you in your studies. Visit www.takshilalearning.com for details. And subscribe

our Social channel for free video lectures.



Follow us on Blogarama



For Class 11 Maths NCERT Solutions – Call us: 8800999280/8800999284