Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - Chapter Summary, Review, and Test - Review Exercises - Page 758: 21

Answer

$420$

Work Step by Step

We are arranging letters, so order is important. Permutations. not all of the $n=7$ letters are distinct: There are 2 T's, 3 O's 1 R and 1 N We use the Permutations of Duplicate Items formula: The number of permutations of $n$ items, where $p$ items are identical, $q$ items are identical, $r$ items are identical, and so on, is $\displaystyle \frac{n!}{p!q!r!\ldots}$ $\displaystyle \frac{7!}{2!3!1!1!}=\frac{7\times[6]\times 5\times 4\times(3\times 2\times 1)}{[2]\times 1\times(3\times 2\times 1)\times 1\times 1}$ ... after reducing $=7\times 3\times 5\times 4$ $=420$
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