Answer
$_{10}C_8=45$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ the given expression, $
_{10}C_8
$ is equivalent to
\begin{align*}\require{cancel}
_{10}C_8&=
\dfrac{10!}{8!\text{ }(10-8)!}
\\\\&=
\dfrac{10!}{8!\text{ }2!}
\\\\&=
\dfrac{10(9)(8!)}{8!\text{ }2!}
\\\\&=
\dfrac{10(9)(\cancel{8!})}{\cancel{8!}\text{ }2!}
\\\\&=
\dfrac{\cancel{10}^5(9)}{\cancel2(1)}
&\text{(divide by $5$)}
\\\\&=
5(9)
\\&=
45
.\end{align*}
Hence, $
_{10}C_8=45
.$
