Answer
$1260$
Work Step by Step
We know that $n!=n\cdot(n-1)\cdot(n-2)...\cdot2\cdot1.$
Hence,
$\require{cancel}
\dfrac{3!\cdot7!}{4!}\\
=\dfrac{3!\cdot(7\cdot6\cdot5\cdot4!)}{4!}\\
=\dfrac{3!\cdot(7\cdot6\cdot5\cdot\cancel{4!})}{\cancel{4!}}\\
=3\cdot2\cdot1\cdot7\cdot6\cdot5\\
=1260$
