The top 2 teams must be from different brackets.
I couldn't understand the question.In the initial competition,8 teams are separated into 4 groups(with 2 teams each) to compete.And it will give 4 winner teams.But how can we choose the top 3 from 4?
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The winner has 8 options of course, then the 2nd place is chosen from the other side of the draw and the 3rd place is from one of the two "bottom" brackets that contributed the losing semifinalists. I'm assuming that's how 3rd place is decided, losing semifinalists play off. – Joffan May 03 '15 at 15:29
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You have a single elimination tournament. Presumably the listing of teams is given. In a standard tournament if the teams are seeded, we have 1 vs 8, 2 vs 7, 3 vs 6 and 4 vs 5 in the first round. Even if they are not seeded, we can use those names for the teams. Then in the second round we have the 1-8 winner play the 4-5 winner and the 2-7 winner play the 3-6 winner and in the third round we have the final. Then the third place team is decided by playing the second round losers.
You have eight choices for the champion. Given the champion, how many choices are there for the loser of the final? Given the first and second place teams, how many choices are there for third?
Ross Millikan
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The second place has 6 possibilities and the third place has 4 posibilities.And the final answer is :8×6×4.Is that correct? – JimfyWinsy May 04 '15 at 05:13
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No. Second only has four, because it has to come from the other bracket from the winner. Multiplying is right. An third has how many? – Ross Millikan May 04 '15 at 05:16
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That is right, because it has to lose either to the winner or second place in the second round. – Ross Millikan May 04 '15 at 14:48