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: 39


$a.\quad 27,600$ $b.\quad 35,152$ $c.\quad 1104$

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. --- $(a)$ $m_{1}=2 \qquad$ ... K or W $m_{2}=25 \qquad$ ... one has been chosen, 25 letters remain $m_{3}=24 \qquad$ ... two have been chosen, 24 letters remain $m_{4}=23 \qquad$ ... three have been chosen, 23 letters remain Total = $2\times 25\times 24\times 23$ = $27,600$ $(b)$ $m_{1}=2 \qquad$ ... K or W $m_{2}=26 \qquad$ ... any letter $m_{3}=26 \qquad$ ... any letter $m_{4}=26 \qquad$ ... any letter Total = $2\times 26^{3}$ = $35,152$ $(c)$ $m_{1}=2 \qquad$ ... K or W $m_{2}=24 \qquad$ ... any letter, except the letter chosen first, and R $m_{3}=23 \qquad$ ... any letter,\ except the letters chosen above, and R $m_{4}=1 \qquad$ ... must be R Total = $2\times 24\times 23\times 1$ = $1104$
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