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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]

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05 Jul 2012, 23:02
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95% (hard)

Question Stats:

40% (03:25) correct 60% (02:05) wrong based on 164 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 _________________ some people are successful, because they have been fortunate enough and some people earn success, because they have been determined..... please press kudos if you like my post.... i am begging for kudos...lol Last edited by Bunuel on 06 Jul 2012, 01:11, edited 2 times in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 33503 Followers: 5930 Kudos [?]: 73465 [5] , given: 9902 Re: In a business school case competition, the top three teams [#permalink] ### Show Tags 06 Jul 2012, 01:14 5 This post received KUDOS Expert's post 3 This post was BOOKMARKED 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]

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01 Jul 2013, 00:59
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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12 Apr 2014, 21:33
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]

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13 Apr 2014, 02:55
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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: 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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Re: In a business school case competition, the top three teams [#permalink]

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14 Apr 2014, 01:47
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 [#permalink]

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22 Oct 2014, 07:10
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 [#permalink]

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22 Oct 2014, 07:58
Expert's post
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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Re: In a business school case competition, the top three teams [#permalink]

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03 Nov 2015, 18:12
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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Re: In a business school case competition, the top three teams   [#permalink] 03 Nov 2015, 18:12
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# In a business school case competition, the top three teams

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