Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 6 - Section 6.4 - Permutations and Combinations - Exercises - Page 437: 15

Answer

$4,950$

Work Step by Step

The number of combinations of 100 items taken 98 at a time : $C(100,98)=\frac{100!}{98!(100-98)!}=\frac{100!}{98!*2!}=\frac{100*99*98*97*...*3*2*1}{98*97*96*,,,*2*1*2*1}=\frac{100*99}{2}=50*99=4,950$
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