# Combinations

Combination is simply a manner of selection of members from a given group. In such kind of selection the order does not make a difference. Also, there is no repetition taken in the concept of combinations. They are assumed without repetition only.

It is denoted by $^nC_k, (C^n_k)$ or most commonly C (n, k), where k is the numbers of members selected and n is the total number of members. Clearly, k $\leq$ n. If $k > n$, then the combination, $C (n, k)$ = 0.

In general permutation and combination go together. But then what is the difference between the two? The difference is that the in a combination the order of arrangement does not matter at all while in permutation order of arrangement matters. Like permutation combination can also be with or without repetition.

In general permutations and combinations are taken along together. Without permutation one can never find a combination.

## Formula

The formula used for finding a given combination is:

$C (n, k)$ = $\frac{n!}{(k! (n – k)!)}$

Here, n is the total number of items o members and k is the number of members or items chosen from total number of given n.
This can also be written as the binomial coefficient (n k) as below:

$\frac{(n (n – 1) (n – 2) … (n – k + 2) (n – k + 1))}{(k (k – 1) (k – 2) … 2 . 1)}$

The formula for permutation in general is:

$P (n, k)$ = $\frac{n!}{(n – k)!}$

So, in the form of permutations the same formula of combination can now be rewritten as follows:

$C (n, k)$ = $\frac{P (n , k)}{k!}$

That is we simply divide our permutation by k!. We all know that k! is the number of repetitions that are made in a given set of elements. This way we are avoiding counting things double or more times.

## Examples

Combination problems are given below:
Example 1: In a lucky draw chits of ten names are out in a box out of which three are to be taken out. Find the number of ways in which those three names can be taken out.

Solution: The possible number of ways for finding three names out of ten from the box is:

C (10, 3) = $\frac{10!}{(3! 7!)}$ = $\frac{10 * 9 * 8 * 7!} {7! * 3 * 2 * 1}$= $\frac{720}{6}$ = 120

So there are 120 different ways of choosing three names out of the ten from the box.

Example 2: Find the values of 14C5, $^{10}C_8$ and C(7, 2).

Solution: We know that the formula for combination, i.e. $C (n, k)$ = $\frac{n!}{(k! (n - k)!)}$
14C5 = $\frac{14!}{(5! (14 -5)!)}$ = $\frac{14!}{(5!\ 9!)}$ = $\frac{14 . 13.12.11.10.9!}{(5.4.3.2.1 .9!)}$  = 2002

$^{10}C_8$ = $\frac{10!}{(8! (10 -8)!)}$ = $\frac{10!}{(8!\ 2!)}$ = $\frac{10.9.8!}{8!.2)}$  = 45

C(7, 2) = $\frac{7!}{(2! (7 -2)!)}$ = $\frac{7!}{(5!\ 2!)}$ = $\frac{7.6.5!}{5!.2)}$ = 21

Example 3: Let us suppose we have 12 adults and 10 kids as an audience of a certain show. Find the number of ways the host can select three persons from the audiences to volunteer. The choice must contain two kids and one adult.

Solution: As order here does not matter so we have:

C (10, 2) * C (12, 1) = [10 * $\frac{9}{2}$] * [$\frac{12}{1}$] = 45 * 12 = 540.

So there are 540 ways in which the host can choose the volunteers containing two kids and an adult.