Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 6 - Section 6.3 - Permutations and Combinations - Exercises: 14

Answer

4851

Work Step by Step

A positive integer less than 100 can be any of: $1, 2, 3, ... , 99$ Therefore, a set of two positive integers less than 100 can be chosen in C(99, 2) ways. [By the definition of C(n, r) ] C(99, 2) = $\frac{99!}{(99-2)!\times 2!}$ = $\frac{99\times98\times97!}{97!\times 2!}$ =$\frac{99\times98}{2!}$ =$\frac{99\times98}{2}$ =
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