Answer
(a) 21
(b) 2520
(c) 720
Work Step by Step
(a) $\frac{7!}{2!5!} = 21$ sequences are possible.
(b) $\frac{7!}{2!1!1!1!1!1!} = 2520$ sequences are possible.
(c) $6! = 720$ sequences are possible.
Applied Statistics and Probability for Engineers, 6th Edition
by Montgomery, Douglas C.; Runger, George C.
Published by
Wiley
ISBN 10:
1118539710
ISBN 13:
978-1-11853-971-2
Chapter 2 - Section 2-1 - Sample Spaces and Events - Exercises - Page 28: 2-43
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