Answer
$_4C_2=6$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ then
\begin{align*}\require{cancel}
_4C_2&=
\dfrac{4!}{2!\text{ }(4-2)!}
\\\\&=
\dfrac{4!}{2!\text{ }2!}
\\\\&=
\dfrac{4(3)(2!)}{2!\text{ }2!}
\\\\&=
\dfrac{4(3)(\cancel{2!})}{2!\text{ }\cancel{2!}}
\\\\&=
\dfrac{4(3)}{2!}
\\\\&=
\dfrac{4(3)}{2(1)}
\\\\&=
\dfrac{\cancel4^2(3)}{\cancel2(1)}
\\\\&=
2(3)
\\&=
6
.\end{align*}
Hence $
_4C_2=6
.$
