Answer
$\frac{n!}{(n-r)!}$
Work Step by Step
$_nP_r=\frac{n!}{(n-r)!}$
First choice: $n$ possibilities
Second choice: $n-1$ possibilities
Third choice: $n-2$ possibilities
....
$r$th choice: $n-(r-1)$ possibilities
Using the Fundamental Counting Principle:
$n(n-1)(n-2)...[n-(r-1)]=n(n-1)(n-2)...[n-(r-1)]·\frac{(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}$
