Binomial Coefficient
$${n \choose k} = \dfrac{n!}{(n-k)!k!}$$
In english, you can think of this statement as saying
If we have $n$ people, how many ways can we choose $k$ people out of the $n$ people?
Derivation
Definitions: Number of subsets of size $k$, of group $n$ people, where order doesn’t matter.
If we had $n$ people and wanna select $k$ of them, here’s how we’d start.
Let’s pick the first person. There are $n$ choices, so our expression looks like this:
$$ n $$
Let’s pick the second person. There are $n-1$ choices left, since we already picked one.
$$ n (n -1) $$
Same thing for the third one
$$ n (n - 1) (n - 2) $$
This generalizes
$$ n (n - 1) (n - 2) \cdots $$
We stop at $(n-k+1)$ because if $k = 1$, we want to stop at n. If $k = 2$, we want to stop at $n-1$, etc.
$$ n (n - 1) (n - 2) \cdots (n - k + 1) $$
This would be correct if we were picking people in a specific order. But in this case, we could’ve selected these people in any order.
We have to divide by $k!$ because there are $k!$ ways to order the people.
$$ \dfrac{n (n - 1) (n - 2) \cdots (n - k + 1)}{k!} $$
It turns out this is the same as
$$ \dfrac{n!}{(n-k)!k!} $$
Note that if $k > n$ then ${n \choose k} = 0$ because it wouldn’t make sense to pick 11 people out of 10 people.