Answer
$6$
Work Step by Step
Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ the given expression, $
_{4}C_{2}
$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{4!}{2!(4-2)!}
\\\\=
\dfrac{4!}{2!2!}
\\\\=
\dfrac{4(3)(2!)}{2(1)(2!)}
\\\\=
\dfrac{\cancel{4}^2(3)(\cancel{2!})}{\cancel{2}(1)(\cancel{2!})}
\\\\=
\dfrac{6}{1}
\\\\=
6
\end{array}
