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In a business school case competition, the top three teams

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In a business school case competition, the top three teams [#permalink] New post 05 Jul 2012, 22:02
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44% (03:18) correct 56% (01:42) wrong based on 78 sessions
In a business school case competition, the top three teams receive cash prizes of $1000, $ 2000, and $ 3000. while remaining teams are not ranked and do not receive any prizes, there are 6 participating teams, named A, B, C, D, E, F. If team A wins one of the prizes, team B wins also one of the prizes. How many outcomes of the competition are possible ?

A. 18
B. 28
C. 36
D. 84
E. 120
[Reveal] Spoiler: OA

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Last edited by Bunuel on 06 Jul 2012, 00:11, edited 2 times in total.
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Re: In a business school case competition, the top three teams [#permalink] New post 06 Jul 2012, 00:14
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alchemist009 wrote:
In a business school case competition, the top three teams receive cash prizes of $1000, $ 2000, and $ 3000. while remaining teams are not ranked and do not receive any prizes, there are 6 participating teams, named A, B, C, D, E, F. If team A wins one of the prizes, team B wins also one of the prizes. How many outcomes of the competition are possible ?

A. 18
B. 28
C. 36
D. 84
E. 120


We are told that "if team A wins one of the prizes, team B wins also one of the prizes". Consider following cases:

A wins one of the prizes, then B must also win one of the prizes, and in this case we can have 4 triplets: {ABC}, {ABD}, {ABE}, {ABF}. Each triplet can be arranged in 3!=6 ways. Hence in the case when A wins one of the prizes 4*6=24 arrangements are possible.

A does NOT win one of the prizes, then three winners must be from other 5 teams. 3 winners out of 5 (B, C, D, E, F) teams can be chosen in C^3_5=10 ways and each case (for example {CDE}) can be arranged in 3!=6 ways, hence in the case when A does NOT win one of the prizes 10*6=60 arrangements are possible.

Total = 24+60 = 84.

Answer: D.

Hope it's clear.
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Re: In a business school case competition, the top three teams [#permalink] New post 30 Jun 2013, 23:59
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Re: In a business school case competition, the top three teams [#permalink] New post 12 Apr 2014, 20:33
Option D.
If we make cases:
I:A doesn't win:5*4*3=60 cases
+
II:A wins:C(3,1)*C(2,1)*C(4,1)=24 cases
Because A could take any one of three prizes
B could take any of the 2 prizes left
And the third leftover prize could be taken by any one of C,D,E,F.
Total=84 cases
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Re: In a business school case competition, the top three teams [#permalink] New post 13 Apr 2014, 01:55
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!



Alternative solution:

If there would have been no constraint the number of possible scenarios were: 6P3 = 6!/3! = 120
Now we need to calculate cases where A is a winner but B is not
So the other two positions can be taken by C, D , E, F . total number of combination = 4C2 = 4!/(2!*2!) = 6
As the three winners can be arranged among themselves in 3! ways, total number of outcomes with A as winner but no B = 6*6 = 36

Hence the number of outcomes which satisfy the constraint in the question = 120 - 36 = 84.

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Re: In a business school case competition, the top three teams [#permalink] New post 14 Apr 2014, 00:47
ind23 wrote:
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!



Alternative solution:

If there would have been no constraint the number of possible scenarios were: 6P3 = 6!/3! = 120
Now we need to calculate cases where A is a winner but B is not
So the other two positions can be taken by C, D , E, F . total number of combination = 4C2 = 4!/(2!*2!) = 6
As the three winners can be arranged among themselves in 3! ways, total number of outcomes with A as winner but no B = 6*6 = 36

Hence the number of outcomes which satisfy the constraint in the question = 120 - 36 = 84.

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Kudos if the answer helped :)


is there any other alternate solution?
Re: In a business school case competition, the top three teams   [#permalink] 14 Apr 2014, 00:47
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