Answer
$56$
Work Step by Step
This involves combination (because order does not matter) where $n = 8$ and $r=3$.
Using the formula $C(n, r)=\dfrac{n!}{(n-r)!\times r!}$ gives:
$C(8,3)
\\= \dfrac{8!}{(8-3)!3!}
\\=\dfrac{8!}{5!\times3!}
\\=\dfrac{8\times7\times6\times5!}{5!\times3!}
\\=\dfrac{8\times7\times6}{3\times2}
\\= 56$
