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: 19

Answer

$1$

Work Step by Step

We know that $C(n,r)=\dfrac{n!}{(n-r)!r!}$, $0!=1$, and $C(n,0)=1$. Hence, $C(15,15)=\dfrac{15!}{(15-15)!\cdot15!}=\dfrac{15!}{0!\cdot15!}=1$
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