Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding: 52

Answer

$4,989,600$

Work Step by Step

Use the theorem for permutations involving $n$ objects that are not distinct, the formula fr which is: $=\dfrac{n!}{n_1!\cdot n_2! \cdot n_3! \cdot \cdot \cdot n_k}$ In the word MATHEMATICS, there are 11 letters in all with $2$ M's, $2 $T's and $2$ A's. Thus, $n_1=2, n_2=2, \text{ and }n_=2$. Using the formula above gives: Number of words that can be formed = $\dfrac{11!}{2!\times2!\times2!} = 4,989,600$
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