Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 11 - Probability and Statistics - Mid-Chapter Quiz - Page 710: 8

Answer

$_{11}P_9=19,958,400$

Work Step by Step

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 .$
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