Answer
$_5C_2=10$
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}
_5C_2&=
\dfrac{5!}{2!\text{ }(5-2)!}
\\\\&=
\dfrac{5!}{2!\text{ }3!}
\\\\&=
\dfrac{5(4)(3!)}{2!\text{ }3!}
\\\\&=
\dfrac{5(4)(\cancel{3!})}{2!\text{ }\cancel{3!}}
\\\\&=
\dfrac{5(4)}{2!}
\\\\&=
\dfrac{5(4)}{2(1)}
\\\\&=
\dfrac{5(\cancel4^2)}{\cancel2(1)}
\\\\&=
5(2)
\\&=
10
.\end{align*}
Hence $
_5C_2=10
.$
