## Algebra 2 (1st Edition)

$504$
This is the same as asking for the number of permutations of $9$ people taken $3$ at a time. $_nP_r=\frac{n!}{(n-r)!}$, hence here: $_9P_3=\frac{9!}{(9-3)!}=\frac{9!}{6!}=9\cdot8\cdot7=504$