, 2014

Solution:

Answer: $2048$

We first group the players by wins, so let $G_{1}$ be the set of all players with the most wins, $G_{2}$ be the set of all players with the second most wins, $\ldots$, $G_{n}$ be the set of all players with the least wins. By the condition in the problem, everyone in group $G_{i}$ must beat everyone in group $G_{j}$ for all $i < j$. Now, consider the mini-tournament consisting of the matches among players inside a single group $G_{i}$. Each must have the same number of wins, say $x_{i}$. But the total number of games is $\binom{\left|G_{i}\right|}{2}$ and each game corresponds to exactly one win, so we must have $\binom{\left|G_{i}\right|}{2} = \left|G_{i}\right| x_{i} \Longrightarrow \left|G_{i}\right| = 2 x_{i} + 1$. Therefore, the number of players in each $G_{i}$ is odd.

We now have $\sum \left|G_{i}\right| = 6$ and all $\left|G_{i}\right|$ are odd, so we can now do casework on the possibilities.

Case 1: $G_{i}$'s have sizes $5$ and $1$. In this case, there are $2$ ways to permute the groups (i.e. either $\left|G_{1}\right| = 5, \left|G_{2}\right| = 1$ or $\left|G_{1}\right| = 1, \left|G_{2}\right| = 5$). There are $6$ ways to distribute the players into the two groups. There are $24$ possible mini-tournaments in the group of size $5$; to prove this, we label the players $p_{1}, \ldots, p_{5}$ and note that each player has $2$ wins. Without loss of generality, let $p_{1}$ beat $p_{2}$ and $p_{3}$, and also without loss of generality let $p_{2}$ beat $p_{3}$. It's easy to verify that there are $2$ possible mini-tournaments, depending on whether $p_{4}$ beats $p_{5}$ or $p_{5}$ beats $p_{4}$. Since there are $\binom{4}{2} \cdot 2 = 12$ ways to pick the two players $p_{1}$ defeats and choose which one beats the other, there are indeed $12 \cdot 2 = 24$ tournaments. Then the total number of possible tournaments in this case is $2 \cdot 6 \cdot 24 = 288$.

Case 2: The sizes are $3, 3$. In this case, there are $\binom{6}{3} = 20$ ways to distribute the players into the groups, and $2$ possible mini-tournaments in either group, so the total here is $20 \cdot 2 \cdot 2 = 80$.

Case 3: The sizes are $3, 1, 1, 1$. In this case, there are $4$ ways to permute the groups, $\binom{6}{3} \cdot 6 = 120$ ways to distribute the players into groups, and $2$ possible mini-tournaments in the group of size $3$, for a total of $4 \cdot 120 \cdot 2 = 960$.

Case 4: The sizes are $1, 1, 1, 1, 1, 1$. There are $720$ ways to distribute the players into groups.

The final answer is $288 + 80 + 960 + 720 = 2048$.