## College Algebra (10th Edition)

$504$
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: $\require{cancel} \dfrac{9!}{6!} \\=\dfrac{9(8)(7)(6)(5)(4)(3)(2)(1)}{6(5)(4)(3)(2)(1)} \\=\dfrac{9(8)(7)\cancel{(6)(5)(4)(3)(2)(1)}}{\cancel{6(5)(4)(3)(2)(1)}} \\=9(8)(7) \\= 504$