 # 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

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!)

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

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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