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In a group of 8 semifinalists, all but 2 will advance to the [#permalink]
31 Aug 2004, 11:44
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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?
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.
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? _________________
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 _________________
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.
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
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.
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.