Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - Chapter Review - Review Exercises - Page 868: 18



Work Step by Step

We know that the number of arrangements of $n$ objects in $r$ slots (where in a slot only $1$ of the $n$ elements can be put) is: $n^r$. If there are $k$ experiments, the first one can be done in $a_1$ ways, the second one in $a_2$ ways... the last one in $a_k$, then there are $a_1\cdot a_2\cdot...\cdot a_k$ ways of doing the experiments together. Hence here, since there are $26-2=24$ letters for the first choice, $9$ digits for the first digit, $10$ digits for the remaining three, the number of possibilities: $24\cdot9\cdot10^3=216000.$
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