11.1. In each of the four volleyball teams, there are six players, including a captain and a setter, and these are different people. In how many ways can a team of six players be formed from these four teams, such that there is at least one player from each team and there must be a pair of captain and setter from at least one team?
Answer: 9720.

Solution. Case 1. Three players, including the captain and the setter, are chosen from one of the teams, and one player is chosen from each of the remaining three teams. The team is chosen in four ways, the third player from it in another four ways, and the three players from the remaining teams in $6 \cdot 6 \cdot 6=216$ ways, resulting in $4 \cdot 4 \cdot 6 \cdot 6 \cdot 6=16 \cdot 216=3456$ possibilities for this case.

Case 2. Two players are chosen from two teams, with at least one team including the captain and the setter, and one player is chosen from each of the remaining two teams. The pair of teams can be chosen in $C_{4}^{2}=6$ ways, and choosing a pair of players from each can be done in $C_{6}^{2} \cdot C_{6}^{2}$ ways. However, we need to exclude $\left(C_{6}^{2}-1\right) \cdot\left(C_{6}^{2}-1\right)$ selections where neither of the two teams includes both the captain and the setter simultaneously, leaving us with $2 \cdot C_{6}^{2}-1=29$ ways. There are $6 \cdot 6=36$ ways to choose one player from each of the remaining two teams, resulting in a total of $6 \cdot 29 \cdot 36=6264$ ways in this case. In total, $6264+3456=9720-$ is the answer to the problem.

Grading criteria. Reducing the solution to cases 1 and 2: 1 point. Considering each of cases 1 and $2: 3$ points.