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by dtweah » Wed May 27, 2009 9:11 am
In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?

(A) 20
(B) 56
(C) 120
(D) 560
(E) 720
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Source: — Problem Solving |

by p2pg » Wed May 27, 2009 9:19 am
Is it D... Not sure if i got it right... 8C6*6C2
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by dumb.doofus » Wed May 27, 2009 9:58 am
On second thoughts..

Well, ultimately in the end, we have three medal winners out of 8..

so 8C3 = 56

Can it be that simple? I doubt though..
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by avenus » Wed May 27, 2009 10:20 am
dumb.doofus wrote:On second thoughts..

Well, ultimately in the end, we have three medal winners out of 8..

so 8C3 = 56

Can it be that simple? I doubt though..
It can... I agree
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by DarkKnight » Wed May 27, 2009 5:45 pm
I would also go with 8C3=56.

What is the OA?
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by vinshu » Wed May 27, 2009 9:23 pm
is it D, 560.

6 out of 8 will advance to final round in 8C6 ways. that is 28 ways.
in turn, 3 out of these 6 will get medals in 6C3 ways which is 20.

Total combinations will be 28*20= 560 ways.
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by dtweah » Thu May 28, 2009 1:31 am
avenus wrote:
dumb.doofus wrote:On second thoughts..

Well, ultimately in the end, we have three medal winners out of 8..

so 8C3 = 56

Can it be that simple? I doubt though..
It can... I agree


OA is 56. Statement of Rounds is just a Red Herring.
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