From a competition archive (like my previous question):
A round-robin chess tournament is held such that each team plays against each other exactly once. Each team loses 10 games and wins 10 games each; there are no draws. How many subsets $\{A,B,C\}$ are there, such that A beats B, B beats C, and C beats A?
The answer given is 385.
My attempt:
As there are 21 teams and each wins 10 teams (or alternatively $\binom{21}2$), we have 210 subsets $\{A,B\}$ such that A beats B. Furthermore, as B also wins 10 times, we have 2100 subsets $\{A,B,C\}$ such that A beats B and B beats C. But how do we apply C beats A?
C wins 10 times and loses 9 times (excluding the loss against B), so there's a roughly $\frac{10}{19}$ chance that C beats A. This gives an approximation of $\frac{10}{19}\cdot2100\approx1105$ subsets. Not taking order into account, we get $\frac{1105}3\approx368$ which is quite close. How do we get the exact value?
Since each team played 20 games against distinct teams, there must be 21 teams. Consider any 3-subset {A,B,C} from all 21 teams. There are $\binom{21}{3} = 1330$ such subsets.
For each 3-subset, it can be the case that one team won 2 games or no team won 2 games.
For the first case, when A beat B and A beat C, there are $\binom{10}{2} = 45$ ways to choose these 2 wins from A's 10 wins, and since this can be the case for any of the 21 teams, there are 45 * 21 = 945 total combinations. For the second case, it must be the case that A beat B, B beat C, and C beat A (since if this is not the case, then some team won 2 games, which is a contradiction). Therefore, this case has 1330 - 945 = 385 combinations since it must account for all the combinations not previously accounted for.