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: 7

Answer

126

Work Step by Step

The number of possible combinations if r items are taken from n items is nCr = $\frac{n!}{r!(n-r)!}$ 9C5 = $\frac{9!}{5!(9-5)!}$ =$\frac{9!}{5!4!}$ =$\frac{9*8*7*6*5!}{5!4!}$ Since 5! is identical in both numerator and denominator so cancel 5! =$\frac{9*8*7*6}{4*3*2*1}$ = 126
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