Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.3 Combinations - Exercise Set 11.3: 21

Answer

0

Work Step by Step

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!},\qquad {}_{n}C_{r}=\frac{n!}{(n-r)!r!}=\frac{{}_{n}P_{r}}{r!}$ --------------- $\displaystyle \frac{{}_{7}P_{3}}{3!}- {}_{7}C_{3}={}_{7}C_{3}-{}_{7}C_{3}=0$
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