Math problem

[BBegemott]

Originally posted by Geoschmo:
I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution