Answer
$56$
Work Step by Step
Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ the given expression, $
_8C_3
$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{8!}{3!(8-3)!}
\\\\=
\dfrac{8!}{3!5!}
\\\\=
\dfrac{8(7)(6)(5!)}{3(2)(1)(5!)}
\\\\=
\dfrac{8(7)(\cancel{6})(\cancel{5!})}{\cancel{3(2)}(1)(\cancel{5!})}
\\\\=
\dfrac{56}{1}
\\\\=
56
\end{array}
