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each of the 6 teams in the league played each of the other

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Eternal Intern
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each of the 6 teams in the league played each of the other [#permalink] New post 08 Jun 2003, 17:19
each of the 6 teams in the league played each of the other teams exactly twice and there were no ties, how many games did the 6 teams play?
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 [#permalink] New post 09 Jun 2003, 10:27
I have the same answer. Just another approach

First team has to play 10 games (6 teams, so 5 games. As they play twice, there are 10 games)
Second team play the same 10 games, but the ones with team one are already counted, so 8 games
Third the same as second team 6 games
Fourth 4 games
Fifth 2 games
Sixth 0

total 10+8+6+4= 30
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 [#permalink] New post 09 Jun 2003, 10:28
Sorry, 10+8+6+4+2 = 30

I forgot the 2
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 [#permalink] New post 14 Mar 2006, 19:50
6C2*2 = 30.

Total number of pairs possible = 6C2.
Number matches between any two paris = 2
So, total matches = 6C2 * 2.

Nice one from the golden grave!
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 [#permalink] New post 14 Mar 2006, 19:51
I too got 30 and the logic is same as that of MBA04's.
tzolkin's approach is simpler. I learnt one more application of combinations today. Thanks tzolkin.
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 [#permalink] New post 14 Mar 2006, 21:26
# of ways = 6C2 * 2 = 30 ways

Use combination because we are interested in grouping and not ordering
  [#permalink] 14 Mar 2006, 21:26
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each of the 6 teams in the league played each of the other

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