Both the permutation and combination deals with computing the number of possible ways to arrange a set of elements. Permutation is an arrangements of different elements in an ordered sequence, whereas combination is grouping the different elements in any order, here order does not matter.

Below are basic permutation and combination formulas:

1. Permutation = n$P_{r}$ = $\frac{n!}{(n-r)!}$.

2. Permutation with repetition = n$^{r}$.

3. Combination = n$C_{r}$ = $\frac{nP_{r}}{n!}$.

4. When permutation is not related, combination = n$C_{r}$ = $\frac{n!}{(n-r)!r!}$.
Where, n = total size of the elements, r = number of permutation.

Below are some of the examples based on permutations and combinations to make you understand properly:

Question 1: There is a competition, which has 15 entities. What is the possibility that only three can win the competition?
Solution:
Given: n = 15, r = 3

Substituting all these values in, 

nP$_{r}$ = $\frac{n!}{(n-r)!}$

15P$_{3}$ = $\frac{15!}{(15-3)!}$

15P$_{3}$ = $\frac{15\times14\times13\times12!}{12!}$

15P$_{3}$ = 2730

Therefore, 2730 permutations of 15 entities are chosen 3 at a time.


Question 2: 5 people came for interview in a company. But only 2 can be selected for the appropriate post. How many ways are there to select them?
Solution:
Given: n = 5, r = 2.

Substituting all these values in, 

nC$_{r}$ = $\frac{n!}{r!(n-r)!}$

5C$_{2}$ = $\frac{5!}{2!(5-2)!}$

5C$_{2}$ = $\frac{5\times4\times3!}{2!3!}$

5C$_{2}$ = 10 

Therefore, 10 combinations of 5 people are chosen 2 at a time.