Combinations without repetition

The combinations without repetition of $n$ elements taken $k$ in $k$ are the different groups of $k$ elements that can be formed by these $n$ elements, so that two groups differ only if they have different elements (that is to say, the order does not matter). They are represented as $C_{n, k}$ .

For example,

Example

Let's consider the set $A = {a, b, c, d, e}$ of $5$ elements. Let's observe first of all that, for example, the groups $a b c$ and $c b a$ are considered to be equal, since as has been said the order does not matter while the elements are the same.

We are going to see what the different combinations without repetition of these $5$ elements are:

Combinations without repetition of $5$ elements taken $1$ at a time: $a$ , $b$ , $c$ , $d$ and $e$ .
Combinations without repetition of $5$ elements taken $2$ at a time: $a b$ , $a c$ , $a d$ , $a e$ , $b c$ , $b d$ , $b e$ , $c d$ , $c e$ and $d e$ .
Combinations without repetition of $5$ elements taken $3$ at a time: $a b c$ , $a b d$ , $a b e$ , $a c d$ , $a c e$ , $a d e$ , $b c d$ , $b c e$ , $b d e$ and $c d e$ .
Combinations without repetition of $5$ elements taken $4$ at a time: $a b c d$ , $a b c e$ , $a b d e$ , $a c d e$ and $b c d e$ .
Combinations without repetition of $5$ elements taken $5$ at a time: The only group of $5$ elements that it is possible to form from the elements of $A$ is $a b c d e$ .

In this example all of the combinations could have been written. However, if $A$ had had many more elements, this would have been much more complicated.

The following formula allows us to know how many combinations without repetition of $n$ elements taken $k$ in $k$ there are: $C_{n, k} = (\binom{n}{k}) = \frac{n!}{k! (n - k)!}$

Example

In the previous example, $n = 5$ . Now, if we want to know how many combinations of $5$ elements, taken $3$ at a time there are, we use the formula and we obtain: $C_{5, 3} = (\binom{5}{3}) = \frac{5!}{3! (5 - 3)!} = 10$ We can check in the previous list that there are $10$ sets of $3$ elements, indeed.