Answer
$_7C_3=35$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ the given expression, $
_7C_3
,$ is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{7!}{3!\text{ }(7-3)!}
\\\\&=
\dfrac{7!}{3!\text{ }4!}
\\\\&=
\dfrac{7(6)(5)(4!)}{3(2)(1)\text{ }4!}
\\\\&=
\dfrac{7(6)(5)(\cancel{4!})}{3(2)(1)\text{ }\cancel{4!}}
\\\\&=
\dfrac{7(\cancel6)(5)}{\cancel3(\cancel2)(1)}
\\\\&=
7(5)
\\&=
35
.\end{align*}
Hence, $
_7C_3=35
.$