Answer
$2\left(_5C_4\right)-_3C_2=7$
Work Step by Step
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ then
\begin{align*}\require{cancel}
_5C_4&=
\dfrac{5!}{4!\text{ }(5-4)!}
\\\\&=
\dfrac{5!}{4!\text{ }1!}
\\\\&=
\dfrac{5!}{4!\text{ }1}
\\\\&=
\dfrac{5!}{4!}
\\\\&=
\dfrac{5(4!)}{4!}
\\\\&=
\dfrac{5(\cancel{4!})}{\cancel{4!}}
\\\\&=
5
.\end{align*}
Using $
_nC_r=\dfrac{n!}{r!\text{ }(n-r)!}
$ or the Combination of $n$ taken $r,$ then
\begin{align*}\require{cancel}
_3C_2&=
\dfrac{3!}{2!\text{ }(3-2)!}
\\\\&=
\dfrac{3!}{2!\text{ }1!}
\\\\&=
\dfrac{3!}{2!}
\\\\&=
\dfrac{3(2!)}{2!}
\\\\&=
\dfrac{3(\cancel{2!})}{\cancel{2!}}
\\\\&=
3
.\end{align*}
Using $_5C_4=5$ and $_3C_2=3,$ the givcen expression, $
2\left(_5C_4\right)-_3C_2
,$ is equivalent to
\begin{align*}
&
2\left(5\right)-3
\\&=
10-3
\\&=
7
.\end{align*}
Hence, $
2\left(_5C_4\right)-_3C_2=7
.$
