Answer
$56$
Work Step by Step
If we want to choose $k$ elements out of $n$, disregarding the order and not allowing repetition, we use combinations:
$_{n}C_k=\frac{n!}{(n-k)!k!}$
Thus we have:
$_{8}C_{3}=\frac{8!}{(8-3)!3!}=\frac{8!}{5!3!}=\frac{8*7*6*5!}{5!*3*2*1}=56$
