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11 players of Team A and 11 players of Team B sit at a round table, the players from the two teams alternate, the goalkeepers sit together, but the captains do not sit together. In how many different ways can they be seated?

Here is my attempt. assume the goalkeepers are sitting on the same chair (since they must be sitting together). There are 20 spots for them. If one of the captain is sitting next to the goalkeeper,(either left or right side, either from team A or team B) there are 2*2*9 ways for the other captain to sit. If captains are not sitting next to the goalkeepers, there are (8*8-(22-6-1)) ways. and the rest of the players have (8!*8!) ways. So totally 2*2*9*(8*8-(22-6-1))*(8!*8!) ways.

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The usual convention about round tables is that arrangements that differ only by a rotation are considered the same.

So we can assume that the goalkeeper from Team A sits in a particular chair. The other goalkeeper has $2$ choices.

Now count the arrangements, without worrying about the restriction on the captains. Fill the seats counterclockwise from the rightmost goalie. The first free chair can be filled in $10$ ways. For every such choice, the next free chair can be filled in $10$ ways. Then the next chair can be filled in $9$ ways, as can the next, and so on, giving a total of $(2)(10!)^2$.

Now we must count the forbidden configurations, in which the two captains (assumed to be non-goalies) sit next to each other.

Start again from the two ways the goalies can sit. The nearest captain (counterclockwise from the right-hand goalie) can be in any one of $19$ chairs. That determines the team membership of that captain, and the position of the other captain. Now count the number of ways to fill in the rest of the positions. This will be $(9!)^2$. So the number of "bad" seatings is $(2)(19)(9!)^2$. Subtract.

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