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In a business school case competition, the top three teams [#permalink]
05 Jul 2012, 22:02

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Difficulty:

85% (hard)

Question Stats:

47% (03:19) correct
53% (01:45) wrong based on 88 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 ?

Re: In a business school case competition, the top three teams [#permalink]
06 Jul 2012, 00:14

1

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Expert's post

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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.

Re: In a business school case competition, the top three teams [#permalink]
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]
13 Apr 2014, 01:55

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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.

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

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

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

is there any other alternate solution?

gmatclubot

Re: In a business school case competition, the top three teams
[#permalink]
14 Apr 2014, 00:47