Why does n choose k always yield a whole number?

Question

I'm just wondering why the n choose k formula

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

always gets you a whole number. I know that obviously there can only be a whole number of "ways" to choose k items in n items, but I'm more interested on how the factorials divide.

Answer 1

Note that $k<n$. So when you write out what $n!$ contains, you will see $k!$ and $(n-k)!$ are included.