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: 12

Answer

$ 132$

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: $\require{cancel} \dfrac{12!}{10!} \\=\dfrac{12(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{10(9)(8)(7)(6)(5)(4)(3)(2)(1)} \\=\dfrac{12(11){(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}}{10(9)(8)(7)(6)(5)(4)(3)(2)(1)} \\=12(11) \\= 132$
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