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: 1

Answer

$20$

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{6!}{3!3!} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{6(5)(4)(3!)}{3!3!} \\\\= \dfrac{6(5)(4)(\cancel{3!})}{3!\cancel{3!}} \\\\= \dfrac{6(5)(4)}{3(2)(1)} \\\\= \dfrac{\cancel{6}(5)(4)}{\cancel{3(2)}(1)} \\\\= \dfrac{(5)(4)}{(1)} \\\\= 20 \end{array}
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