Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 9 - Counting and Probability - Exercise Set 9.2 - Page 537: 17

Answer

a. 9000 integers b. 4500 odd integers c. 4536 integers with distinct digits d. 2240 odd integers with distinct digits e. 50.4% probability for a random 4-digit integer with distinct digits 24.9% for random 4-digit odd integer with distinct digits

Work Step by Step

a. 9*10*10*10 = (9999 – 1000) + 1 = 9000 integers b. 9*10*10*5 = ((9999 – 1000) + 1)/2 = 4500 odd integers c. 9*9*8*7 = 4536 integers with distinct digits d. 5*8*8*7 = 2240 odd integers with distinct digits 5 (4th digit is odd) * 8 (1st digit not zero and not matching the last digit) * 8 (2nd digit not matching last and 1st and 4th digits) * 7 (3rd digit not matching 1st, 2nd, 4th digits) = 5*8*8*7 = 2240 e. 4536/9000 = 504/1000 = 63/125 = 50.4% probability for a random 4-digit integer with distinct digits 2240/9000 = 224/900 = 56/225 = 24.9% for random 4-digit odd integer with distinct digits
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.