Permutation and Combination, both are used to count the ways of possibilities. In simple words, Permutation involves only with arrangements of the objects in various ways that they can be ordered whereas Combination is all about selecting the objects and order does not matter.The formulas for Permutation and Combination are stated as below :

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Below are the problems based on permutations and combinations :

### Solved Examples

**Question 1: **Find the number of permutations and combinations if n = 15 and r = 3 ?

** Solution: **
**Step 1: **as

**Step 2: **

Given n = 15 and r = 3

Permutation Formula: P(n, r) = $\frac{n!}{(n - r)!}$ = $\frac{15!}{(15 - 3)!}$ = $\frac{15!}{(12!)}$ = $\frac{15 \times 14 \times 13 \times 12!}{(12!)}$ = 2730

Combination Formula: C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{15!}{3!(15 - 3)!}$ = $\frac{15!}{3!(12!)}$ = $\frac{15 \times 14 \times 13 \times 12}{3!(12!)}$ = 455

**Question 2: **Find the number of permutations and combinations if n = 12 and r = 2 ?

** Solution: **

Given n = 12 and r = 2

Permutation Formula: P(n, r) = $\frac{n!}{(n - r)!}$ = $\frac{12!}{(12 - 2)!}$ = $\frac{12!}{(10!)}$ = $\frac{12 \times 11 \times 10!}{(10!)}$ = 132

Combination Formula: C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{12!}{2!(12 - 2)!}$ = $\frac{12!}{2!(10!)}$ = $\frac{12 \times 11 \times 10!}{2!(10!)}$ = 66

Given n = 15 and r = 3

Permutation Formula: P(n, r) = $\frac{n!}{(n - r)!}$ = $\frac{15!}{(15 - 3)!}$ = $\frac{15!}{(12!)}$ = $\frac{15 \times 14 \times 13 \times 12!}{(12!)}$ = 2730

Combination Formula: C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{15!}{3!(15 - 3)!}$ = $\frac{15!}{3!(12!)}$ = $\frac{15 \times 14 \times 13 \times 12}{3!(12!)}$ = 455

Given n = 12 and r = 2

Permutation Formula: P(n, r) = $\frac{n!}{(n - r)!}$ = $\frac{12!}{(12 - 2)!}$ = $\frac{12!}{(10!)}$ = $\frac{12 \times 11 \times 10!}{(10!)}$ = 132

Combination Formula: C(n, r) = $\frac{n!}{r!(n - r)!}$ = $\frac{12!}{2!(12 - 2)!}$ = $\frac{12!}{2!(10!)}$ = $\frac{12 \times 11 \times 10!}{2!(10!)}$ = 66

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