Thinking Mathematically (6th Edition)

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: 7

Answer

The statement is false. An example of a true statement: Even though BET and BEE are both three letter words, the number of permutations for each of them differ. BET has 3!=6 permutations. BEE has $\displaystyle \frac{3!}{2!}$=3 permutations.

Work Step by Step

-------------------- For the word BET, all three letters are different, so it is true that the number of permutations is $3!=6.$ For the word BEE, there is a group of duplicates (two Es), for which the formula is different, $\displaystyle \frac{3!}{2!}=\frac{6}{2}=3$ The statement is false. An example of a true statement: Even though BET and BEE are three letter words, the number of permutations for each of them differ. BET has 3!=6 permutations. BEE has $\displaystyle \frac{3!}{2!}$=3 permutations.
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