Answer
0
Work Step by Step
${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!},\qquad {}_{n}C_{r}=\frac{n!}{(n-r)!r!}=\frac{{}_{n}P_{r}}{r!}$
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$\displaystyle \frac{{}_{20}P_{2}}{2!}- {}_{20}C_{2}= {}_{20}C_{2}- {}_{20}C_{2}=0$
Thinking Mathematically (6th Edition)
by Blitzer, Robert F.
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 - Page 707: 22
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