# Find the number of combinations

Suppose we are selecting a few items from a large number of distinct items and you are trying to find out how many different ways this can be done. You are trying to find the number of combinations.

Suppose you want to buy 2 cars from a set of 4 distinct choices. Suppose the set is {Honda, Toyota, Mercedes, BMW}

How many ways can you select two cars from this set?

Here are the different ways to do this.

(Honda, Toyota)  (Honda, Mercedes)  (Honda, BMW)  (Toyota, Mercedes) (Toyota, BMW)  (Mercedes, BMW)

Each of the possible selections in this list is called a combination. Therefore, the number of combinations is 6.

It is important to notice that the order in which the selections are made is not important when dealing with combinations.

For instance, the combination (Honda, Mercedes) is the same as (Mercedes, Honda). Both of these arrangement represent 1 combination.

Combination notation

Combinations give the number of ways r items can be selected from n items.

The notation that we use to denote the number of combinations is  C(n, r) or nCr

You can also use the following notation. Notice that the two notation above and the one below can be read as "the number of combinations of n items selected r at a time."

$${n \choose r}$$

n is the total number of items

r is the number of items selected from the total numbers of items

## Number of combinations formula

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

Let us revisit our previous example about selecting 2 cars from a set of 4 choices. Use the formula above to find the number of combinations

In this case, n = 4 and r = 2

$${n \choose r} = {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!(2)!}$$
$${n \choose r} = {4 \choose 2} = \frac{4×3×2!}{2!(2)!} = \frac{4×3}{2!} = \frac{12}{2×1!} = \frac{12}{2×1}$$
$${n \choose r} = {4 \choose 2} = \frac{12}{2} = 6$$

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