Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 5 - Section 5.5 - Assess Your Understanding - Applying the Concepts - Page 306: 45

Answer

In $59,280$ ways.

Work Step by Step

The order in which the cars finish the race matters and one car can not finish in two different positions. Permutation of 40 distinct cars taken 3 at a time: $_3P_{40}=\frac{40!}{(40-3)!}=\frac{40!}{37!}$ But, $40!=40\times39\times38\times(37\times36\times...\times3\times2\times1)=4\times39\times38\times37!$ $_3P_{40}=\frac{40\times39\times38\times37!}{37!}=40\times39\times38=59,280$
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