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

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In a group of 8 semifinalists, all but 2 will advance to the [#permalink] New post 01 Dec 2004, 11:29
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In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 wil be awarded medals then how many groups of medal winners are possible?
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 [#permalink] New post 01 Dec 2004, 16:42
C(8,6) * C(6,3)

Does all but 2 means all of them except 2 ?
It would be then 8-2 = 6....right ?
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 [#permalink] New post 01 Dec 2004, 23:48
The way to read the problem under discussion is that => from 8 it gets to 6 and from 6 it gets to 3. We will take the same flow for easy logic, hence it is 6 out of 8 (which is 8C6) and 3 out of 6 (which is 6C3) => totalling to 8C6*6C3 ways
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 [#permalink] New post 02 Dec 2004, 00:44
Thats how I did it too... but the answer's 8C3...which is 56.

The explanation given is that since 3 people are finally chosen from 8 people, we can take it as 8C3 and NOT 8C6*6C3 ( this is what even I did)... :?

Can we have some explanations please?? Why is it 8C3??
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PS- Medal winners [#permalink] New post 02 Dec 2004, 04:24
I get 8C3 = 56

I think that if you do 8C6 * 6C3 you are counting many groups more than once:

Let’s say you have A, B ,C , D, E, F, G, H, and A and B do not qualify for the final: we have 6C3 = 20 possibilities (eg, CDE, CDF).
Now we leave out G and H, you get another 20 possibilities.
You can do this 8C6 = 28 times, but many groups get repeated. In our example CDE will be a group that appears when A and B don’t qualify and also appear when G and H don’t qualify or when A and G don’t qualify. That’s why we are counting all groups many times.

It does not matter that only 6 got to the final, we have 8 elements to play with and we have to put them into groups of 3. Going from the semifinal to the final is not adding any restriction (eg A never qualifies). And if there was a restriction the answer could have never be more than 8C3
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Re: PS- Medal winners [#permalink] New post 02 Dec 2004, 12:05
artabro wrote:
I get 8C3 = 56

I think that if you do 8C6 * 6C3 you are counting many groups more than once:

Let’s say you have A, B ,C , D, E, F, G, H, and A and B do not qualify for the final: we have 6C3 = 20 possibilities (eg, CDE, CDF).
Now we leave out G and H, you get another 20 possibilities.
You can do this 8C6 = 28 times, but many groups get repeated. In our example CDE will be a group that appears when A and B don’t qualify and also appear when G and H don’t qualify or when A and G don’t qualify. That’s why we are counting all groups many times.

It does not matter that only 6 got to the final, we have 8 elements to play with and we have to put them into groups of 3. Going from the semifinal to the final is not adding any restriction (eg A never qualifies). And if there was a restriction the answer could have never be more than 8C3


Artabro,

Thanks a lot for trying to explain this to me.. But I couldnt understand it... :cry: :cry:
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 [#permalink] New post 02 Dec 2004, 18:15
It makes sense in the hind sight.

here instead of asking how many ways can you pick 3 out of 8, the author is twisting the question by explaining how he would select. Since we don't really know or care about the selection process we can safely say 3 can be picked from 8 in 8C3 ways.
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 [#permalink] New post 02 Dec 2004, 20:37
praveen_rao7 wrote:
It makes sense in the hind sight.

here instead of asking how many ways can you pick 3 out of 8, the author is twisting the question by explaining how he would select. Since we don't really know or care about the selection process we can safely say 3 can be picked from 8 in 8C3 ways.



Isn't it 6C3 = 20. 2 out of 8 will not make it to the final, and we are looking for the combination of medals in the final round.

What is the OA.
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 [#permalink] New post 02 Dec 2004, 21:10
Please read Ruhi's post. OA is 8C3
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