Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 47

Answer

$132,860$

Work Step by Step

Order matters so the question is involves permutation. We know that the number of permutations for $n$ objects taken $r$ at a time is given by the formula $P(n,r)=\dfrac{n!}{(n-r)!}.$ Here we have $n=365$ and $r=2$. Hence, $P(365,2)=\dfrac{365!}{(365-2)!}=\dfrac{365\cdot364\cdot363!}{363!}=365\cdot364=132860$.
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