Answer
$_7C_3=35$
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}
_7C_3&=
\dfrac{7!}{3!\text{ }(7-3)!}
\\\\&=
\dfrac{7!}{3!\text{ }4!}
\\\\&=
\dfrac{7(6)(5)(4!)}{3!\text{ }4!}
\\\\&=
\dfrac{7(6)(5)(\cancel{4!})}{3!\text{ }\cancel{4!}}
\\\\&=
\dfrac{7(6)(5)}{3!\text{ }}
\\\\&=
\dfrac{7(6)(5)}{3(2)(1)}
\\\\&=
\dfrac{7(\cancel6)(5)}{\cancel3(\cancel2)(1)}
\\\\&=
7(5)
\\&=
35
.\end{align*}
Hence, $
_7C_3=35
.$
