Answer
$1,260$
Work Step by Step
RECALL:
(1) For any natural number $n$,
$n! = n(n-1)(n-2)(n-3)$
(2) $0!=1$
Use the definition in (1) above to obtain:
$\require{cancel}
\dfrac{3!7!}{4!}
\\=\dfrac{(3\cdot2\cdot1)(7(6)(5)(4)(3)(2)(1))}{(4)(3)(2)(1)}
\\=\dfrac{(3\cdot2\cdot1)(7(6)(5)\cancel{(4)(3)(2)(1))}}{\cancel{(4)(3)(2)(1)}}
\\=\dfrac{6 \cdot (42(5))}{1}
\\= \dfrac{6\cdot210}{1}
\\=1,260$
