Answer
$_7C_5=21$
Work Step by Step
Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ then the value of the given expression, $
_7C_5
,$ is
\begin{align*}\require{cancel}
&
\dfrac{7!}{5!(7-5)!}
\\\\&=
\dfrac{7!}{5!2!}
\\\\&=
\dfrac{7(6)(5!)}{5!2(1)}
\\\\&=
\dfrac{7(6)(\cancel{5!})}{\cancel{5!}2(1)}
\\\\&=
\dfrac{42}{2}
\\\\&=
21
.\end{align*}
Hence, the given expression evaluates to $
_7C_5=21
.$
