College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Test - Page 791: 6

Answer

$720$

Work Step by Step

$_{n}P_k=\frac{n!}{(n-k)!}$ Hence here it is $_{10}P_3=\frac{10!}{(10-3)!}=10\cdot9\cdot8=720$
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