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

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Director
Joined: 03 Jul 2003
Posts: 652
In a group of 8 semifinalists, all but 2 will advance to the [#permalink]

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31 Aug 2004, 12:44
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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 will be awarded medals then how many groups of medals winners are possible?
Intern
Joined: 10 Aug 2004
Posts: 40

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31 Aug 2004, 13:14
I don't think you need to include the 8C6. If only 6 advance to the final round, the information that the competition began with 8 is only meant to try and throw one off. I think the answer is straight 6C3...20.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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31 Aug 2004, 13:24
Let's say that the 2 left out in the first round were John and Marc. When forming the groups of 3 to award medals, we will be ignoring all the combinations with those 2 when forming the groups of 3 with only the 6 finalists are we not?
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Best Regards,

Paul

Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US

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31 Aug 2004, 13:34
Paul is right.

We're only looking for possibilities. John and Marc are part of these possibilities.

Without John and Marc, yes, we would only have 20 possibilities.
Intern
Joined: 03 Aug 2004
Posts: 26

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31 Aug 2004, 13:56
does the order of finish matter, or just that 3 medal?

if it doesn't, then I agree with Paul

Last edited by lovely on 31 Aug 2004, 13:58, edited 1 time in total.
Manager
Joined: 02 Apr 2004
Posts: 222
Location: Utrecht

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31 Aug 2004, 13:58

I follow questions like these already for a while, but since last week I am active within the club. You guys rock with questions like this!

After research and instructions from the forum I finally can make an answer on my own.

8!/6!2! * 6!/3!3! which is 560

Regards,

Alex
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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31 Aug 2004, 14:16
lovely, the question is about "how many groups of medals winners are possible?". So the order of medal is not important.

Alex, combination problems are problem types that are lacking in every GMAT book I could find. There has been a resurgence of those types of problems in the GMAT and the best source I found was on this website. When I first started, I knew nothing about combinations and although I still get many combination questions wrong I learned a great deal from this website. Congratulation on your understanding of this concept now
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Best Regards,

Paul

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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31 Aug 2004, 21:33
It has to be 8C6 * 6C3

We need to find the number of possibilites going through to the last round. Note, the possibilites does not care about positioning. -- combinations, so use 8C6

In the last rond, we have 3 winners, but this time, we need to care about the order, who's first, who's second and who's third. We can pick only 3 from this pool of 6, so it must be done using permutations -- 6C3

For 1 combination of 6, there are 6C3 permutations,

For for 8C6 combinations, there are 8C6*6C3 possibilites.

Hope this is clear enough.
Intern
Joined: 08 Jan 2004
Posts: 31
Location: Seattle

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02 Sep 2004, 22:23
I don't think the intermediate step is relevant here - we have to choose 3 out of 8 - does not matter how we chose 3 - we might have reduced it to a set of 6 first and then chosen 3 or reduced it to a set of 4 and then chosen 3 - the answer is the same.
8C3 = 56
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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02 Sep 2004, 23:40
No ! To get to the next round, we have to eliminate 2 contestants. So we'll end up with 8C6 possible groups of 6 finalists. From this pool, we have another 3 winning medals. So the possible ways to get groups of 3 is 6C3. But we are asked total possibilites of winning medals, to get this we need to multiply 8C6 by 6C3. Why ?

Assuming:
Team 1 (is one of 8C6 possibilites) --> Has 6C3 possibilites to win medals
Team 2 (another of 8C6 possibilites) -> Has 6C3 possibilites to win mdeals

So 8C3 teams each having 6C3 possibilites is = 8C3*6C3.

It's like

Group 1 has 2 ways to win a contest
Group 2 has also 2 ways win the contest
Together, they have 4 ways between themselves to win the same contest.
Director
Joined: 20 Jul 2004
Posts: 592

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02 Sep 2004, 23:47
Seattle_76 wrote:
I don't think the intermediate step is relevant here - we have to choose 3 out of 8 - does not matter how we chose 3 - we might have reduced it to a set of 6 first and then chosen 3 or reduced it to a set of 4 and then chosen 3 - the answer is the same.
8C3 = 56

Completely agree with this. A loser is a loser irrespective of when he loses

Problem is simple in that it doesn't confuse us with first medal, second medal and so forth. Bit it only raises questions with level 1 rejection and level two rejection.

Paul wrote:
Let's say that the 2 left out in the first round were John and Marc. When forming the groups of 3 to award medals, we will be ignoring all the combinations with those 2 when forming the groups of 3 with only the 6 finalists are we not?

With names not mentioned, J amd M could be anyone. Everyone gets the same treatment and possibility at this early stage.
Director
Joined: 03 Jul 2003
Posts: 652

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05 Sep 2004, 05:36
The OA is 56, which I consider incorrect.

I got 560. My method is same as Paul's.
05 Sep 2004, 05:36
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