## Algebra 2 Common Core

$_{11}P_9=19,958,400$
Using $_nP_r=\dfrac{n!}{(n-r)!}$ or the Permutation of $n$ taken $r,$ the given expression, $_{11}P_9 ,$ is equivalent to \begin{align*}\require{cancel} & \dfrac{11!}{(11-9)!} \\\\&= \dfrac{11!}{2!} \\\\&= \dfrac{11(10(9)(8)(7)(6)(5)(4)(3)(2!)}{2!} \\\\&= \dfrac{11(10(9)(8)(7)(6)(5)(4)(3)(\cancel{2!})}{\cancel{2!}} \\\\&= 11(10(9)(8)(7)(6)(5)(4)(3) \\&= 19,958,400 .\end{align*} Hence, $_{11}P_9=19,958,400 .$