College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.1 - Sequences - 9.1 Assess Your Understanding: 14

Answer

$806,400$

Work Step by Step

RECALL: (1) For any natural number $n$, $n! = n(n-1)(n-2)(n-3)$ (2) $0!=1$ Use the definition in (1) above to obtain: $\require{cancel} \dfrac{5!8!}{3!} \\=\dfrac{(5\cdot 4 \cdot 3\cdot2\cdot1)(8(7)(6)(5)(4)(3)(2)(1))}{(3)(2)(1)} \\=\dfrac{(5\cdot 4 \cancel{\cdot 3\cdot2\cdot1})(8(7)(6)(5)(4)(3)(2)(1))}{\cancel{(3)(2)(1)}} \\=\dfrac{5 \cdot (40320)}{1} \\= \dfrac{806,400}{1} \\=806,400$
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