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: 13

Answer

$3360.$

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. --- $m_{1}=16$ ... 16 desserts to choose from, for 1st place $m_{2}=15$ ... of the 15 remaining, one will have won 2nd place $m_{3}=14$ ... of the 14 remaining, one will have won 3rd place Total = $16\times 15\times 14$ = $3360.$ Another approach: Permutations, r=3 taken from n=16: $P(n,r)=\displaystyle \frac{n!}{(n-r)!}=\frac{16\times 15\times 14\times 13!}{(16-3)!}$=$16\times 15\times 14$
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