Answer
$10$
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{5!}{2!3!}
,$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{5(4)(3!)}{2(1)(3!)}
\\\\=
\dfrac{5(4)(\cancel{3!})}{2(1)(\cancel{3!})}
\\\\=
\dfrac{5(\cancel{4}^2)}{\cancel{2}(1)}
\\\\=
\dfrac{10}{1}
\\\\=
10
\end{array}
