## Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning

# Chapter 9 - Counting and Probability - Exercise Set 9.2 - Page 537: 15

#### Answer

a) Possible combinations = 30^{3} = 27000 b) P(30,3) = \frac{30}{30-3} = 30*29*28 = 24360

#### Work Step by Step

a) Since repetition is allowed, we can choose 3 numbers in the following ways first number can be chosen in 30 ways (any number between 1-30). second number can be chosen in 30 ways (any number between 1-30). third number can be chosen in 30 days(any number between 1-30). So, total combinations = 30*30*30 b) Since repetition is not allowed, we can choose 3 numbers in the following ways first number can be chosen in 30 ways (any number between 1-30). second number can be chosen in 29 ways (any number between 1-30 except the first number). third number can be chosen in 28 days(any number between 1-30 except first and second number). So, total combinations = 30*29*28

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