Answer
$P(n,r)=\dfrac{n!}{(n-r)!}$
Work Step by Step
The first one can be chosen out of $n$ objects, the second one out of $n-1$... the $r$th one out of $n-r+1$.
According to the Multiplication Principle, if we have $p$ selections for the first choice, $q$ selections for the second choice, then we have $p\cdot q$ different ways of selections and similarly for more than $2$ selections.
Hence,
$P(n,r)=n(n-1)(n-2)...(n-r+1)=\dfrac{n!}{(n-r)!}.$
