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In a business school case competition, the top three teams
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Updated on: 06 Jul 2012, 00:11
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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 ? A. 18 B. 28 C. 36 D. 84 E. 120
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Originally posted by alchemist009 on 05 Jul 2012, 22:02.
Last edited by Bunuel on 06 Jul 2012, 00:11, edited 2 times in total.
Edited the question.




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Re: In a business school case competition, the top three teams
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06 Jul 2012, 00:14
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
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30 Jun 2013, 23:59



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Re: In a business school case competition, the top three teams
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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



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Re: In a business school case competition, the top three teams
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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: 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



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Re: In a business school case competition, the top three teams
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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 is there any other alternate solution?



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Re: In a business school case competition, the top three teams
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22 Oct 2014, 06:10
I understand the explanation, however I'm confused about an "assumption."
It says if A wins, then B also wins. From this I assumed that if A didn't win, B didn't win either? Why is this wrong in the context of the wording?
I just assumed it like this because in DS questions, when it has a conditional (if ...), that is usually true.



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Re: In a business school case competition, the top three teams
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22 Oct 2014, 06:58
intheend14 wrote: I understand the explanation, however I'm confused about an "assumption."
It says if A wins, then B also wins. From this I assumed that if A didn't win, B didn't win either? Why is this wrong in the context of the wording?
I just assumed it like this because in DS questions, when it has a conditional (if ...), that is usually true. Don't get your analogy with DS question... Anyways, if A wins, B wins does not mean if B wins, A wins.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: In a business school case competition, the top three teams
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21 Apr 2018, 22:28
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