College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 9, Counting and Probability - Section 9.1 - Counting - 9.1 Exercises: 16

Answer

$156,849$

Work Step by Step

RECALL: $C(n, r) = \dfrac{n!}{r!(n-r)!}$ Use the formula above to obtain: $C(99, 3) = \dfrac{99!}{3!(99-3)!} \\C(99, 3)) = \dfrac{99!}{3!(96!!)} \\C(99, 3) = \dfrac{99(98)(97)(96!)}{3\cdot2\cdot1 \cdot 96!}$ Cancel the common factors to obtain: $\require{cancel} \\C(99, 3) = \dfrac{\cancel{99}33\cancel{(98}49)(97)\cancel{(96!)}}{\cancel{3}\cdot\cancel{2}\cdot1 \cdot \cancel{96!}} \\C(99, 3) = 156,849$
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