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: 38

Answer

$30,030$ ways

Work Step by Step

Committee members are chosen with no importance of order of choice, so we deal with combinations. We have a sequence of selections in which we choose 1. ... 2 out of a group of 5 professors ... in ${}_{5}C_{2}$ ways 2. ... 10 out of a group of 15 students... in ${}_{15}C_{10}$ ways By the Fundamental Counting Principle, Total ways= ${}_{7}C_{4}\cdot {}_{7}C_{5}$ ${}_{5}C_{2}=\displaystyle \frac{5!}{(5-2)!2!}=\frac{5\times 4}{1\times 2}=10$ ${}_{15}C_{10}=\displaystyle \frac{15!}{(15-10)!10!}$ $=\displaystyle \frac{15\times 14\times 13\times 12\times 11}{1\times 2\times 3\times 4\times 5}=3003$ Total = $10\times 3003$ = $30,030$ ways $30,030$ ways
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