## Precalculus (10th Edition)

$35$
The order of the students doesn't matter here, so the situation involves combination. We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$ . Here we have $n=7$ and $r=4$, thus, $C(7,4)=\dfrac{7!}{(7-4)!\cdot4!}=\dfrac{7\cdot6\cdot5\cdot4!}{3\cdot2\cdot1\cdot4!}=\dfrac{7\cdot6\cdot5}{6}=35$.