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: 27

Answer

$\displaystyle \frac{3}{68}$

Work Step by Step

${}_{n}C_{r}=\displaystyle \frac{n!}{(n-r)!r!}$ -------- ${}_{4}C_{2}=\displaystyle \frac{4!}{(4-2)!2!}=\frac{4\times 3}{1\times 2}=6$ ${}_{6}C_{1}=\displaystyle \frac{6!}{(6-1)!1!}=6$ ${}_{18}C_{3}=\displaystyle \frac{18!}{(18-3)!3!}=\frac{18\times 17\times 16}{1\times 2\times 3}$ $=3\times 17\times 16=816$ $\displaystyle \frac{{}_{4}C_{2}\cdot {}_{6}C_{1}}{{}_{18}C_{3}}=\frac{6\times 6}{816}=\frac{3}{68}$
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