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12 Jul 2014, 23:52
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (01:52) correct
43%
(02:16)
wrong
based on 437
sessions
History
Date
Time
Result
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A certain football coach allows his quarterback to call three plays in a row. If the quarterback can choose from 3 passing plays, 4 running plays, and 2 trick plays, how many different arrangements of plays can be run, if the same play can’t be run more than once?
A. 24
B. 125
C. 243
D. 504
E. 729
A. 24
B. 125
C. 243
D. 504
E. 729
13 Jul 2014, 07:21
Kudos
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Bunuel : is this approach correct.
Number of ways to choose the first play: 9
Second play: 8
Third:7
9*8*7= 504
Number of ways to choose the first play: 9
Second play: 8
Third:7
9*8*7= 504
General Discussion
13 Jul 2014, 05:18
Kudos
Bookmarks
anceer
P=3, R=4, and T=2.
There are following cases possible:
1. Different kinf of plays:
PRT (Any combination) = 3*4*2*3!=144.
2. Two same kind of plays and one from other kind of play:
PPR (Any combination. Notice that the two P's here must be different) = \(C^2_3*4*3!=72\);
PPT = \(C^2_3*2*3!=36\);
RRP = \(C^2_4*3*3!=108\);
RRT = \(C^2_4*2*3!=72\);
TTP = \(C^2_2*3*3!=18\);
TTR = \(C^2_2*4*3!=24\);
3. All three plays are of the same kind:
PPP = C^3_3*3!=6
RRR = C^3_4*3!=24
TTT is not possible because we have only two T's.
Total = 144 + 72 + 36 + 108 + 72 + 18 + 24 + 6 + 24 = 504.
Answer: D.
