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: 23

Answer

$\displaystyle \frac{3}{4}$

Work Step by Step

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!}$ --------------- ${}_{3}P_{2}=\displaystyle \frac{3!}{(3-2)!}=3\times 2=6$ ${}_{4}P_{3}=\displaystyle \frac{4!}{(4-3)!}=4\times 3\times 2=24$ $1-\displaystyle \frac{ {}_{3}P_{2}}{{}_{4}P_{3}}=1-\frac{6}{24}=1-\frac{1}{4}=\frac{3}{4}$
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