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 - Page 701: 53


$105$ 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 $\displaystyle \frac{n!}{p!q!r!\ldots}$ ---------------- 5,446,666 contains n=7 digits, of which there are 4 duplicate 6s ... p=4, 2 duplicate 4s ... q=2 a single 5 ..... in all n=7 (checking) ... So, the total number of distinct permutations is $\displaystyle \frac{7!}{4!2!}=\frac{7\times[6]\times 5\times(4\times 3\times 2\times 1)}{(4\times 3\times 2\times 1)\times[2]}$ $=7\times 3\times 5$ $=105$ ways
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.