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: 51

Answer

$90,720$

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 this example, there are $2$ O's and $2$ C's. Thus, we have $N_1=2$ and $n_2=2$. Therefore, Number of words that can be formed = $\dfrac{9!}{2!\times2!} = 90720$.
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