Answer
$20$
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{6!}{3!3!}
,$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{6(5)(4)(3!)}{3!3!}
\\\\=
\dfrac{6(5)(4)(\cancel{3!})}{3!\cancel{3!}}
\\\\=
\dfrac{6(5)(4)}{3(2)(1)}
\\\\=
\dfrac{\cancel{6}(5)(4)}{\cancel{3(2)}(1)}
\\\\=
\dfrac{(5)(4)}{(1)}
\\\\=
20
\end{array}
