Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 10 Counting Methods and Probability - 10.1 Apply the counting Principles and Permutations - 10.1 Exercises - Problem Solving - Page 688: 65



Work Step by Step

This is the same as asking for the number of permutations of $9$ people taken $3$ at a time. $_nP_r=\frac{n!}{(n-r)!}$, hence here: $_9P_3=\frac{9!}{(9-3)!}=\frac{9!}{6!}=9\cdot8\cdot7=504$
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