If ${ }^{n+2}{C}_3{=}^{n+3}{P}_2-20$ then $n$ is equal to:

$\frac{1}{6}(n+2)(n+1)n=(n+3)(n+2)-20$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{n}^3+3n^2+2n=6\left(n^2+5n+6\right)-120\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{n}^3-3n^2-28n+84=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(n-3)\left(n^2-28\right)=0\end{array}$

As $n$ is a positive integer we get $n=3$