Answer
$_9C_8=9$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ the given expression, $
_9C_8
,$ is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{9!}{8!\text{ }(9-8)!}
\\\\&=
\dfrac{9!}{8!\text{ }1!}
\\\\&=
\dfrac{9(8!)}{8!\text{ }1}
\\\\&=
\dfrac{9(\cancel{8!})}{\cancel{8!}\text{ }1}
\\\\&=
9
.\end{align*}
Hence, $
_9C_8=9
.$
