Answer
$_5C_3=10$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ the given expression, $
_5C_3
$ is equivalent to
\begin{align*}\require{cancel}
_5C_3&=
\dfrac{5!}{3!\text{ }(5-3)!}
\\\\&=
\dfrac{5!}{3!\text{ }2!}
\\\\&=
\dfrac{5(4)(3!)}{3!\text{ }2!}
\\\\&=
\dfrac{5(4)(\cancel{3!})}{\cancel{3!}\text{ }2!}
\\\\&=
\dfrac{5(4)}{2(1)}
\\\\&=
\dfrac{5(\cancel4^2)}{\cancel2(1)}
&\text{(divide by $2$)}
\\\\&=
5(2)
\\&=
10
.\end{align*}
Hence, $
_5C_3=10
.$
