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: 28

Answer

$\displaystyle \frac{21}{44}$

Work Step by Step

${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ -------- ${}_{5}C_{1}=\displaystyle \frac{5!}{(5-1)!1!}=5$ ${}_{7}C_{2}=\displaystyle \frac{7!}{(7-2)!2!}=\frac{7\times 6}{1\times 2}=21$ ${}_{12}C_{3}=\displaystyle \frac{12!}{(12-3)!3!}=\frac{12\times 11\times 10}{1\times 2\times 3}$ $=2\times 11\times 10=220$ $\displaystyle \frac{{}_{5}C_{1}\cdot {}_{7}C_{2}}{{}_{12}C_{3}}=\frac{5\times 21}{220}=\frac{21}{44}$
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