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: 4



Work Step by Step

Using $n!=n(n-1)(n-2)\cdot\cdot\cdot(3)(2)(1)$ or the factorial of a number, the given expression, $ \dfrac{8!}{5!3!} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{8(7)(6)(5!)}{5!(3)(2)(1)} \\\\= \dfrac{8(7)(\cancel{6})(\cancel{5!})}{\cancel{5!}\cancel{(3)(2)}(1)} \\\\= \dfrac{8(7)}{1} \\\\= 56 \end{array}
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