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: 17

Answer

$35$

Work Step by Step

We know that $C(n,r)=\frac{n!}{(n-r)!r!}$and $C(n,0)=1$. Hence, $C(7,4)=\dfrac{7!}{(7-4)!\cdot4!}=\dfrac{7\cdot6\cdot5\cdot4!}{3!\cdot4!}=\dfrac{7\cdot6\cdot5}{3\cdot2\cdot1}=35$
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