Answer
$504$
Work Step by Step
We know that $n!=n\cdot(n-1)\cdot (n-2)...\cdot 3\cdot2\cdot 1$
Use this definition to obtain:
$$\displaystyle
\require{cancel} \dfrac{9!}{6!} =\dfrac{9\cdot8\cdot7\cdot6!}{6!}\\
=\dfrac{9\cdot8\cdot7\cdot\cancel{6!}}{\cancel{6!}}\\
=9\cdot8\cdot7\\
=504$$