Thinking Mathematically (6th Edition)

Published by Pearson

Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Exercise Set 11.2 - Page 701: 54

Answer

$630$ ways

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 $\displaystyle \frac{n!}{p!q!r!\ldots}$ ---------------- 5,432,435 contains n=7 digits, of which there are 2 duplicate 5s ... p=2, 2 duplicate 4s ... q=2 2 duplicate 3s ... r=2 a single 2 ..... in all n=7 (checking) ... So, the total number of distinct permutations is $\displaystyle \frac{7!}{2!2!2!}=\frac{7\times 6\times 5\times(4)\times 3\times[2]\times 1}{(2\times 2)\times[2]}$ $=7\times 6\times 5\times 3$ $=630$ ways

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