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
Thinking Mathematically (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0321867327
ISBN 13:
978-0-32186-732-2
Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Concept and Vocabulary Check - Page 700: 5
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