Proof
If \(r \gt n\) or \(r \lt 0 \) then this is not possible and so the result would be no permuatations. Otherwise, apply the multiplication principle r times noting that there are n choices for the first selection, n-1 choices for the second selection, and with n-r+1 choices for the rth selection. This gives
\begin{align*}
_nP_r & = n(n-1) ... (n-r+1)\\
& = n(n-1) ... (n-r+1)\frac{(n-r)!}{(n-r)!}\\
& = \frac{n(n-1) ... (n-r+1)(n-r)!}{(n-r)!}\\
& = \frac{n!}{(n-r)!}
\end{align*}