Answer
$56$
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{8!}{5!3!}
,$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{8(7)(6)(5!)}{5!(3)(2)(1)}
\\\\=
\dfrac{8(7)(\cancel{6})(\cancel{5!})}{\cancel{5!}\cancel{(3)(2)}(1)}
\\\\=
\dfrac{8(7)}{1}
\\\\=
56
\end{array}