Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.2 Permutations and Combinations - 13.2 Asses Your Understanding - Page 854: 2

Answer

false

Work Step by Step

We know that $n!=n\cdot(n-1)\cdot...\cdot2\cdot1.$ Hence, $(n+1)n!=(n+1)!\\n!=\frac{(n+1)!}{n+1}$ Thus, the given statement is false.
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