Answer
$35$
Work Step by Step
Using $n!=n(n-1)(n-2)\cdot\cdot\cdot(3)(2)(1)$ or the factorial of a number, the given expression, $
\dfrac{7!}{3!4!}
,$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{7(6)(5)(\cancel{4!})}{3(2)(1)(\cancel{4!})}
\\\\=
\dfrac{7(\cancel{6})(5)}{\cancel{3(2)}(1)}
\\\\=
\dfrac{35}{1}
\\\\=
35
\end{array}
