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: 84

Answer

$6840$

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 $\dfrac{20!}{(20-3)}=\dfrac{20 !}{17 !}$ $=\dfrac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9\cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{17 !}$ $=20 \cdot 19 \cdot 18$ $=6840$
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