Precalculus (10th Edition)

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

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 855: 30


The combinations: $123,124,125,126,134,135,136,145,146,156$ $234,235,236,245,246,256,345,346,356,456$ $C(6,3)=20$

Work Step by Step

The combinations: $123,124,125,126,134,135,136,145,146,156$ $234,235,236,245,246,256,345,346,356,456$ We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ Hence $C(6,3)=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}=20$.
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