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: 2

Answer

$10$

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{5!}{2!3!} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{5(4)(3!)}{2(1)(3!)} \\\\= \dfrac{5(4)(\cancel{3!})}{2(1)(\cancel{3!})} \\\\= \dfrac{5(\cancel{4}^2)}{\cancel{2}(1)} \\\\= \dfrac{10}{1} \\\\= 10 \end{array}
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