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805+ (Hard)|   Combinations|      
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alchemist009
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In a business school case competition, the top three teams Updated on: 06 Jul 2012, 01:11
Originally posted by alchemist009 on 05 Jul 2012, 23:02.
Last edited by Bunuel on 06 Jul 2012, 01:11, edited 2 times in total.
Edited the question.
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95% (hard)

Question Stats:

39% (02:35) correct 61% (02:30) wrong based on 304 sessions

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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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D
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Bunuel
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In a business school case competition, the top three teams 06 Jul 2012, 01:14
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alchemist009
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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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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In a business school case competition, the top three teams 01 Jul 2013, 00:59
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
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In a business school case competition, the top three teams 12 Apr 2014, 21:33
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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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In a business school case competition, the top three teams 13 Apr 2014, 02:55
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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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.

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In a business school case competition, the top three teams 14 Apr 2014, 01:47
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Bunuel
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.

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is there any other alternate solution?
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In a business school case competition, the top three teams 22 Oct 2014, 07:10
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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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In a business school case competition, the top three teams 22 Oct 2014, 07:58
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intheend14
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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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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In a business school case competition, the top three teams 17 Feb 2025, 00:45
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