Answer
$126$
Work Step by Step
Since $\left(\begin{array}{c} n\\ r\end{array} \right)=\dfrac{n!}{r!(n-r)!}$, then, $\left(\begin{array}{c}
9 \\ 5
\end{array} \right)$ is equal to
\begin{array}{l}\require{cancel}
\dfrac{9!}{5!(9-5)!}
\\\\=
\dfrac{9!}{5!4!}
\\\\=
126
\end{array}