## Algebra and Trigonometry 10th Edition

$\frac{n!}{(n-r)!}$
$_nP_r=\frac{n!}{(n-r)!}$ 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: $n(n-1)(n-2)...[n-(r-1)]=n(n-1)(n-2)...[n-(r-1)]·\frac{(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}$