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: 14

Answer

$252$

Work Step by Step

RECALL: $C(n, r) = \dfrac{n!}{r!(n-r)!}$ Use the formula above to obtain: $C(10, ,5) = \dfrac{10!}{5!(10-5)!} \\C(10, 5) = \dfrac{10!}{5!(5!)} \\C(10, 5) = \dfrac{10(9)(8)(7)(6)(5!)}{5\cdot 4\cdot3\cdot2\cdot1 \cdot 5!}$ Cancel the common factors to obtain: $\require{cancel} \\C(10, 5) = \dfrac{10(9)(8)(7)\cancel{(6)}\cancel{(5!)}}{5\cdot 4\cdot\cancel{3\cdot2\cdot1} \cdot \cancel{5!}} \\C(10, 5)= \dfrac{\cancel{10}2 \cdot 9 \cdot \cancel{8}2 \cdot 7}{\cancel{5} \cdot \cancel{4}} \\C(10, 5) = 252$
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