Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 767: 55

Answer

A)35152 B)913952

Work Step by Step

A)Since the first letter is already fixed, we need to find possible combinations of the last 3 letters: There are 2 choices (W or K)for the 1st letter There are 26 choices for the second. There are 26 choices for the 3rd. There are 26 choices for the 4th. The number of call signs: 2 $\times$26$\times$26$\times$26=35152. There are 35152 possible combinations. B)The first letter is already fixed so we need to find the possible combination or the last 4 letters: There are 2 choices (W or K )for the 1st letter. There are 26 choices for the 2nd letter. There are 26 choices for the 3rd letter. There are 26 choices for the 4th letter. There are 26 choices for the 5th letter. The number of call signs is 2 $\times$26$ \times$26 $\times$ 26 $\times $26=913952 There are 913952 possible combinations
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