Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 6 - Section 6.5 - Generalized Permutations and Combinations - Exercises - Page 433: 31

Answer

$83160$

Work Step by Step

In "ABRACADABRA", there are 5 A's, 2 B's, 1 C, 1 D, and 2 R's, for a total of 11 letters. So we are trying to calculate the number of permutations of 11 letters, with the 5 A's indistinguishable from each other, 2 B's indistinguishable from each other, and 2 R's indistinguishable from each other. By Theorem 3 on page 428, the number of permutations is: $\frac{11!}{5!2!2!}$ $=\frac{1*2*3*4*5*6*7*8*9*10*11}{1*2*3*4*5*1*2*1*2}$ $=\frac{6*7*8*9*10*11}{2*2}$ $=83160$
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