College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 9, Counting and Probability - Section 9.1 - Counting - 9.1 Exercises - Page 656: 29

Answer

$\boxed {158184000}$

Work Step by Step

We can repeat letters so we can choose 9 characters for the first blank, 26 for the next 3, as there are 26 letters in the alphabet, and then 10 choices for the last 3. $10^3*26^3*9$ = $\boxed {158184000}$
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