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27 Apr 2020, 02:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:27) correct
68%
(02:25)
wrong
based on 85
sessions
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Eight teams participate in a tournament and each team plays exactly one match with every other team. The winner gets 1 point and the loser gets 0 while there are no ties. Find the total possible ranking outcome of tournaments such that A always gets more points than B and C both at the end of the tournament.
A) 48
B) 56
C) 13440
D) 20160
E) 40320
A) 48
B) 56
C) 13440
D) 20160
E) 40320
Source: https://www.GMATinsight.com
27 Apr 2020, 03:49
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GMATinsight
Solution:
- • Ranking of A must be above B and C.
- o Out of 8 spots we need to select 3 spots for A, B, and C.
- Number of ways of selecting 3 spots = 8C3
- Number of ways in which these 3 teams can be ranked = 8C3*2!
- Total number of total possible ranking = 8C3*2!*5! = 13,440.
12 Jul 2024, 08:04
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The 8 teams can be ranked without restriction:
8! ways
If the ranking requirement were:
ABC
Then 8! would need to be divided by 3! = 6 to account for the different arrangements of A,B and C.
But because there is no requirement that B defeat C or visa versa, 6 as the divisor is reduced by a factor of 2, so that the final result is:
8!/3 = 8*7*6*5*4*2 =
56*40*6 = 2240*6 = 13440
Posted from my mobile device
8! ways
If the ranking requirement were:
ABC
Then 8! would need to be divided by 3! = 6 to account for the different arrangements of A,B and C.
But because there is no requirement that B defeat C or visa versa, 6 as the divisor is reduced by a factor of 2, so that the final result is:
8!/3 = 8*7*6*5*4*2 =
56*40*6 = 2240*6 = 13440
Posted from my mobile device
