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: 3



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{7!}{3!4!} ,$ evaluates to \begin{array}{l}\require{cancel} =\dfrac{7(6)(5)(\cancel{4!})}{3(2)(1)(\cancel{4!})} \\\\= \dfrac{7(\cancel{6})(5)}{\cancel{3(2)}(1)} \\\\= \dfrac{35}{1} \\\\= 35 \end{array}
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