## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$56$
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: $C(8,3) \\= \dfrac{8!}{(8-3)!3!} \\=\dfrac{8!}{5!\times3!} \\=\dfrac{8\times7\times6\times5!}{5!\times3!} \\=\dfrac{8\times7\times6}{3\times2} \\= 56$