6.1. Four chess players - Ivanov, Petrov, Vasiliev, and Kuznetsov - played a round-robin tournament (each played one game against each other). A win earns 1 point, a draw earns 0.5 points each. It turned out that the player who took first place had 3 points, and the player who took last place had 0.5 points. How many variants of point distribution among the named chess players exist, if some of them could have scored the same number of points? (For example, the variants where Ivanov has 3 points and Petrov has 0.5, and where Petrov has 3 points and Ivanov has 0.5, are considered different!)
Answer: 36

Solution. First, let's determine what quantities of points the four participants can have under the given conditions. Since the one who took first place has 3 points, he won against everyone, and the one who took second place could not have scored more than 2, as he lost to the first. There were a total of 6 games in the tournament, which means 6 points were distributed (regardless of the outcome of the game, 1 point is always at stake), so the second and third places together have $6-3-0.5=2.5$ points. Therefore, they could have scored 2 and 0.5 or 1.5 and 1 point. In the first case, we need to find the number of ways to distribute the points $3 ; 2 ; 0.5$ and 0.5 among the four given chess players. This can be done in $4 \cdot 3=12$ ways, as any of the four can be in first place, and any of the remaining three can be in second place, with no further choice. In the second case, there will be $4 \cdot 3 \cdot 2=24$ ways, since here the points are different for all. In the end, we get $12+24=36$ ways to distribute the points.