Answer
$3\left(_3P_2\right)+_3P_1=21$
Work Step by Step
Using $
_nP_r=\dfrac{n!}{(n-r)!}
$ or the Permutation of $n$ taken $r,$ then
\begin{align*}\require{cancel}
_3P_2&=
\dfrac{3!}{(3-2)!}
\\\\&=
\dfrac{3!}{1!}
\\\\&=
\dfrac{3(2)(1)}{1}
\\\\&=
6
.\end{align*}
Using $
_nP_r=\dfrac{n!}{(n-r)!}
$ or the Permutation of $n$ taken $r,$ then
\begin{align*}\require{cancel}
_3P_1&=
\dfrac{3!}{(3-1)!}
\\\\&=
\dfrac{3!}{2!}
\\\\&=
\dfrac{3(2!)}{2!}
\\\\&=
\dfrac{3(\cancel{2!})}{\cancel{2!}}
\\\\&=
3
.\end{align*}
Using $_3P_2=6$ and $_3P_1=3,$ the given expression, $
3\left(_3P_2\right)+_3P_1
,$ is equivalent to
\begin{align*}
&
3\left(6\right)+3
\\&=
18+3
\\&=
21.\end{align*}
Hence, $
3\left(_3P_2\right)+_3P_1=21
.$
