Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Test - Page 1127: 26



Work Step by Step

There are a total of $^{15}C_{6}$ possible ways to choose $6$ numbers for the winning lottery combination and this can be our denominator. We know that $ n!=1 \cdot 2 \cdot 3 .....(n-1)n $ Thus, we have $\dfrac{50!}{^{15}C_{6}}=\dfrac{50 !}{\dfrac{15! }{(15-6)!}} $ $= \dfrac{50}{5005}$
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