## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Concept and Vocabulary Check: 5

#### Answer

False. $8!=8(7)(6)(5)(4)(3)(2)(1)$

#### Work Step by Step

See: Factorial Notation, page 696. $n!=n(n-1)(n-2)\cdots(3)(2)(1)$ and $0!=1$ -------------- The statement is false. To make it true, change each addition sign to multiplication

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