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: 13



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: $\require{cancel} \dfrac{3!7!}{4!} \\=\dfrac{(3\cdot2\cdot1)(7(6)(5)(4)(3)(2)(1))}{(4)(3)(2)(1)} \\=\dfrac{(3\cdot2\cdot1)(7(6)(5)\cancel{(4)(3)(2)(1))}}{\cancel{(4)(3)(2)(1)}} \\=\dfrac{6 \cdot (42(5))}{1} \\= \dfrac{6\cdot210}{1} \\=1,260$
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