## Precalculus (10th Edition)

$56$
The order of the professors doesn't matter here, so the situation involves combination. We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$. Thus, with $n=8$ and $r=3$, then $C(8,3)=\dfrac{8!}{(8-3)!\cdot3!}=\dfrac{8\cdot7\cdot6\cdot5!}{5!\cdot3\cdot2\cdot1}=\dfrac{8\cdot7\cdot6}{6}=56$.