Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

00:00

A

B

C

D

E

Difficulty:

(N/A)

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.