College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 7 - Section 7.4 - The Binomial Theorem - 7.4 Exercises - Page 664: 8

Answer

$84$

Work Step by Step

Using $\left( \array{n\\r} \right)=\dfrac{n!}{r!(n-r)!},$ the given expression, $\left( \array{ 9\\3 } \right)$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{9!}{3!(9-3)!} \\\\= \dfrac{9!}{3!6!} \\\\= \dfrac{9(8)(7)(6!)}{3(2)(1)(6!)} \\\\= \dfrac{\cancel{9}^3(\cancel{8}^4)(7)(\cancel{6!})}{\cancel{3}(\cancel{2})(1)(\cancel{6!})} \\\\= \dfrac{84}{1} \\\\= 84 \end{array}
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