[Tournaments and tournament tables] $[$ Combinations and arrangements $]$

12 chess players played a round-robin tournament. Then each of them wrote 12 lists. The first list contained only themselves, the $(k+1)$-th list contained those who were in the $k$-th list and those they had defeated. It turned out that for each chess player, the 12th list differed from the 11th. How many draws were there?
Consider a directed graph where the vertices are chess players, and arrows point from the winner to the loser. The condition means that for each chess player, there is another player who can be reached by exactly 11 arrows (this, in particular, means that from each chess player, one can reach any other player). Consider such a path: $A_{1}$ won against $A_{2}, A_{2}$ won against $A_{3}, \ldots, A_{11}$ won against $A_{12}$. Note that $A_{i}(1<i<12)$ could not have won against $A_{1}$ (otherwise, from $A_{2}$, one could reach each player in no more than 10 arrows). But someone must have won against $A_{1}$ (otherwise, it would be impossible to reach $A_{1}$ at all), so it must be $A_{12}$. Similarly, we show that in the resulting cycle, each player could only win against the next one.

Therefore, there are 12 decisive games, and the number of draws is $-12 \cdot 11: 2-12=54$.

## Answer

54 draws.