Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.6 Basics of Counting Theory - 11.6 Exercises - Page 1061: 9

Answer

$160$ ways.

Work Step by Step

Apply the Fundamental Principle of Counting If $n$ independent events occur, with $m_{1}$ ways for event 1 to occur, $m_{2}$ ways for event 2 to occur, $\ldots$ and $m_{n}$ ways for event $n$ to occur, then there are $m_{1}\cdot m_{2}\cdot\cdots\cdot m_{n}$ different ways for all $n$ events to occur. --- $m_{1}=2$ $m_{2}=20$ $m_{3}=4$ Total = $2\times 20\times 4$ = $160$ ways.
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