Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 12 - Counting and Probability - Chapter Test - Page 889: 7

Answer

$462$

Work Step by Step

Recall the combination formula: $C(n,k)=\frac{n!}{(n-k)!k!}$ We choose $5$ out of $11$ or $C(11,5)$. Thus: $C(11,5)=\require{cancel} \frac{11!}{(11-5)!5!}=\frac{11\cdot10\cdot9\cdot8\cdot7\cdot6!}{6!5!}=\frac{11\cdot10\cdot9\cdot8\cdot7\cdot\cancel{6!}}{\cancel{6!}5!}=\frac{11\cdot10\cdot9\cdot8\cdot7}{5!}=462$
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