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: 15

Answer

$C(100, 1) = 100$

Work Step by Step

RECALL: $C(n, r) = \dfrac{n!}{r!(n-r)!}$ Use the formula above to obtain: $C(100, 1) = \dfrac{100!}{1!(100-1)!} \\C(100, 1)) = \dfrac{100!}{1!(99!!)} \\C(100, 1) = \dfrac{100!(1)}{99!} \\C(100, 1) = \dfrac{100!}{99!} \\C(100, 1) = \dfrac{100(99!)}{99!} $ Cancel the common factors to obtain: $\require{cancel} \\C(100, 1) = \dfrac{100\cancel{(99!)}}{\cancel{99!}} \\C(100, 1) = 100$
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