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# In a group of 8 semifinalists, all but 2 will advance to the

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Senior Manager
Joined: 20 Feb 2006
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In a group of 8 semifinalists, all but 2 will advance to the [#permalink]

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28 Jun 2006, 04:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

43. In a group of 8 semifinalists, all but 2 will advance to the final round. In the final round only the top 3 will be awarded medals, how many groups of medal winners are possible?

(A) 20
(B) 56
(C) 120
(D) 560
(E) 720

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SVP
Joined: 30 Mar 2006
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28 Jun 2006, 04:44
D

Number of ways six semifinalists can be chosen = 8C6 = 28
Number of ways three medal winners can be chosen = 6C3 = 20
Total combination possible = 28* 20 = 560

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Senior Manager
Joined: 07 Jul 2005
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Location: Sunnyvale, CA

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28 Jun 2006, 09:25
Another one for (D) = 560
#ways to select finalists = 8C6 = 28
#ways to select winners = 6C3 = 20

Total = 28 * 20 = 560
(D)

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CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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28 Jun 2006, 10:04
I think midele step is irrelevent.
It should be B
8C3 = 56

Lets take an example. There are 4 people and only two will advance in the next round. Out of those two only one will be awarded a medal. How many combinations of medal winners are possible. 4C1 i.e. 4. Not 4C2 * 2C1 = 6 * 2 = 12

If the question asks how many paths are there to go to the winning group then it should be 4C2*2C1 = 12

Am I missing something????
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Senior Manager
Joined: 20 Feb 2006
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28 Jun 2006, 10:18
ps_dahiya wrote:
I think midele step is irrelevent.
It should be B
8C3 = 56

Lets take an example. There are 4 people and only two will advance in the next round. Out of those two only one will be awarded a medal. How many combinations of medal winners are possible. 4C1 i.e. 4. Not 4C2 * 2C1 = 6 * 2 = 12

If the question asks how many paths are there to go to the winning group then it should be 4C2*2C1 = 12

Am I missing something????

You are right.. ..Answer is 8C3.. And that's teh trick part..

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Manager
Joined: 26 Jun 2006
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28 Jun 2006, 10:25
I got (D) 560 too. The question clearly asks "how many groups of medal winners are POSSIBLE," so all possible combination of people making/not making the final group are important and should be counted. So, 28*20=560

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Manager
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28 Jun 2006, 10:40
I understand why I got it wrong. I overanalyzed this one. 56 is a simpler, one-step solution

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Director
Joined: 06 May 2006
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28 Jun 2006, 10:47
Fell into the trap!!

Great question!
_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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28 Jun 2006, 10:47
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