## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Test - Page 1126: 7

#### Answer

$120$

#### Work Step by Step

We know that $n!=1 \cdot 2 \cdot 3 .....(n-1)n$ Thus, we have $\dfrac{10!}{(10-3)!3!}$ $= \dfrac{10 \cdot 9\cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\times 3 \cdot 2 \cdot 1 }$ $=120$

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