In a group of 8 semifinalists, all but 2 will advance to the : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 00:35

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 78 [0], given: 0

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

### Show Tags

01 Dec 2004, 11:29
00:00

Difficulty:

(N/A)

Question Stats:

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

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?
_________________

Let's get it right!!!!

Manager
Joined: 30 Jul 2004
Posts: 101
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

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 ?
Director
Joined: 16 Jun 2004
Posts: 893
Followers: 3

Kudos [?]: 59 [0], given: 0

### Show Tags

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
Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 78 [0], given: 0

### Show Tags

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??
_________________

Let's get it right!!!!

Manager
Joined: 15 Jul 2004
Posts: 75
Location: London
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

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:

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
Senior Manager
Joined: 19 Oct 2004
Posts: 317
Location: Missouri, USA
Followers: 1

Kudos [?]: 78 [0], given: 0

### Show Tags

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...
_________________

Let's get it right!!!!

Director
Joined: 25 Jan 2004
Posts: 728
Location: Milwaukee
Followers: 3

Kudos [?]: 23 [0], given: 0

### Show Tags

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.
_________________

Praveen

Manager
Joined: 28 Nov 2004
Posts: 76
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

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.
Director
Joined: 25 Jan 2004
Posts: 728
Location: Milwaukee
Followers: 3

Kudos [?]: 23 [0], given: 0

### Show Tags

02 Dec 2004, 21:10
_________________

Praveen

02 Dec 2004, 21:10
Display posts from previous: Sort by