Answer
$504$
Work Step by Step
RECALL:
(1) For any natural number $n$,
$n! = n(n-1)(n-2)(n-3)...(3)(2)(1)$
(2) $0!=1$
Use the definition in (1) above to obtain:
$\require{cancel}
\dfrac{9!}{6!}
\\=\dfrac{9(8)(7)(6)(5)(4)(3)(2)(1)}{6(5)(4)(3)(2)(1)}
\\=\dfrac{9(8)(7)\cancel{(6)(5)(4)(3)(2)(1)}}{\cancel{6(5)(4)(3)(2)(1)}}
\\=9(8)(7)
\\= 504$