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 - Exercise Set 11.2: 51

Answer

$831,600$

Work Step by Step

See Permutations of Duplicate Items, page 699. The number of permutations of $n$ items, where $p$ items are identical, $q$ items are identical, $r$ items are identical, and so on, is $\displaystyle \frac{n!}{p!q!r!\ldots}$ ---------------- TALLAHASSEE contains n=11 letters, of which there are 3 duplicate A's ... p=3, 2 duplicate L's ... q=2 2 duplicate S's .... r=2 2 duplicate E's ... s=2 a single T and a single H ..... in all n=11 (checking) ... So, the total number of distinct permutations is $\displaystyle \frac{11!}{3!2!2!2!}=$ $\dfrac{11\times 10\times 9\times 8\times 7\times(6)\times 5\times(4)\times 3\times(2)\times 1}{(6)\times(2\times 2)\times(2)}$ $=11\times 10\times 9\times 8\times 7\times 5\times 3$ $=831,600$
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