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: 22



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 $C(n,r)=\frac{n(n-1)(n-2)...(n-r+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ a) Hence here $C(5,1)C(8,3)=\frac{5}{1}\frac{8\cdot7\cdot6}{3\cdot2\cdot1}=5\cdot56=280$. b)Hence here $C(5,2)C(8,2)=\frac{5\cdot4}{2\cdot1}\frac{8\cdot7}{2\cdot1}=10\cdot28=280$. c)Exactly 1 man:$280$, exactly 2 men:$280$, exactly $3$ men: $C(5,3)C(8,1)=\frac{5\cdot4\cdot3}{3\cdot2\cdot1}\frac{8}{1}=10\cdot8=80$. Hence in total: $280+280+80=640.$ (The commitee can't have 4 men because it has to have at least 1 woman.)
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