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: 49

Answer

There are $303,600$ possible outcomes.

Work Step by Step

The order in which the balls are placed matters and no ball can be selected twice. Permutation of 25 distinct balls taken 4 at a time: $_{25}P_{4}=\frac{25!}{(25-4)!}=\frac{25!}{21!}$ But, $25!=25\times24\times23\times22\times(21\times20\times19\times ... \times3\times2\times1)=25\times24\times23\times22\times21!$ $_{25}P_{4}=\frac{25\times24\times23\times22\times21!}{21!}=25\times24\times23\times22=303,600$
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