College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 10 - Test - Page 709: 11


See below.

Work Step by Step

According to the Multiplication Principle, if we have $p$ selections for the first choice, $q$ selections for the second choice, then we have $p\cdot q$ different ways of selections and so on for more choices... Here we have $26-3=23$ possibilities for the third character, $26$ possibilities for the first two characters and $10$ possibilities for the four numbers, thus the number of all possibilities is $23\cdot26^2\cdot10^4=155480000$
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