Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1050: 85

Answer

$24,804$

Work Step by Step

Write the correct form as below: We know that $n!=1 \cdot 2 \cdot 3 .....(n-1)n$ Thus, we have $\dfrac{54!}{(54-3)}=\dfrac{54 !}{51!}$ $=\dfrac{54 \cdot 53 \cdot 52 \cdot 51 !}{51 !}$ $=54 \cdot 53 \cdot 52$ $=24,804$
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