## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and the Binomial Theorem - Guided Practice for Examples 1, 2, and 3 - Page 691: 5

#### Answer

$5400$

#### Work Step by Step

We can choose $3$ out of the $10$ tragedies and $2$ of the $10$ histories, thus the number of possible sets of plays: $_{10}C_3\cdot_{10}C_2=\frac{10!}{7!3!}\frac{10!}{8!2!}=\frac{10\cdot9\cdot8\cdot7!}{7!3!}\frac{10\cdot9\cdot8!}{8!2!}=120\cdot45=5400$

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