Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - Chapter Review - Review Exercises - Page 868: 21


a)$381024$. b)$1260$.

Work Step by Step

If we have two groups with $n$ and $m$ objects and we choose $r(r\leq n)$ and $s(s\leq m)$ objects from the two groups where the order doesn't matter then the number of combinations is: $C(n,r)C(m,s).$ (Similarly for more than $2$ groups.) We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ a) Hence here $C(9,4)C(9,3)C(9,2)=\frac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1}\frac{9\cdot8\cdot7}{3\cdot2\cdot1}\frac{9\cdot8}{2\cdot1}=126\cdot84\cdot36=381024$. b)Hence here $C(9,4)C(5,3)C(2,2)$ is what we are looking for (because after the first committee chose $4$ people there are only $9-4=5$ left for the second committee) $C(9,4)C(5,3)C(2,2)=\frac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1}\frac{5\cdot4\cdot3}{3\cdot2\cdot1}\frac{2\cdot1}{2\cdot1}=126\cdot10\cdot1=1260$.
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