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In a group of 8 semifinalists, all but 2 will advance to the [#permalink]
01 Dec 2004, 11:29

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Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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? _________________

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

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

Letâ€™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 donâ€™t qualify and also appear when G and H donâ€™t qualify or when A and G donâ€™t qualify. Thatâ€™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

Re: PS- Medal winners [#permalink]
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... _________________

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. _________________

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.

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