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: 11



Work Step by Step

RECALL: (1) For any natural number $n$, $n! = n(n-1)(n-2)(n-3)...(3)(2)(1)$ (2) $0!=1$ Use the definition in (1) above to obtain: $\require{cancel} \dfrac{9!}{6!} \\=\dfrac{9(8)(7)(6)(5)(4)(3)(2)(1)}{6(5)(4)(3)(2)(1)} \\=\dfrac{9(8)(7)\cancel{(6)(5)(4)(3)(2)(1)}}{\cancel{6(5)(4)(3)(2)(1)}} \\=9(8)(7) \\= 504$
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