Answer
$84$
Work Step by Step
Using $_nC_r=\dfrac{n!}{r!(n-r)!},$ the given expression, $
_{9}C_{3}
$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{9!}{3!(9-3)!}
\\\\=
\dfrac{9!}{3!6!}
\\\\=
\dfrac{9(8)(7)(6!)}{3(2)(1)(6!)}
\\\\=
\dfrac{\cancel{9}^3(\cancel{8}^4)(7)(\cancel{6!})}{\cancel{3}(\cancel{2})(1)(\cancel{6!})}
\\\\=
\dfrac{84}{1}
\\\\=
84
\end{array}
