Algebra 2 Common Core

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

Chapter 11 - Probability and Statistics - 11-2 Probability - Practice and Problem-Solving Exercises - Page 687: 43

Answer

$_7P_4=840$

Work Step by Step

Using $ _nP_r=\dfrac{n!}{(n-r)!} $ or the Permutation of $n$ taken $r,$ the given expression, $ _7P_4 $ is equivalent to \begin{align*}\require{cancel} _7P_4&= \dfrac{7!}{(7-4)!} \\\\&= \dfrac{7!}{3!} \\\\&= \dfrac{7(6)(5)(4)(\cancel{3!})}{\cancel{3!}} \\\\&= 7(6)(5)(4) \\&= 840 .\end{align*} Hence, $ _7P_4=840 .$
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