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In a business school case competition, the top three teams [#permalink]

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05 Jul 2012, 22:02

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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 ?

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.

Re: In a business school case competition, the top three teams [#permalink]

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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

Re: In a business school case competition, the top three teams [#permalink]

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13 Apr 2014, 01:55

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This post received KUDOS

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: 6P\(3\) = 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.

------------------------------------------ Kudos if the answer helped

Re: In a business school case competition, the top three teams [#permalink]

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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: 6P\(3\) = 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.

------------------------------------------ Kudos if the answer helped

Re: In a business school case competition, the top three teams [#permalink]

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03 Nov 2015, 17:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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