Answer
5330
Work Step by Step
This involves a combination since the order does not matter. We know that:
$_nC_r =\frac{_nP_r}{r!} = \frac{n!}{(n-r)!r!}$
$_{16}C_4 = \frac{16!}{(16-4)!4!}$
$\frac{ 16!}{4!12!}$
We also know that:
$x! = x(x-1)(x-2)...(1)$
Thus, we have:
5330