A team is represented by its division and a number, so C5 would be team 5 in division C. It's assumed that a game is a match between 2 teams in the same division. In division B each team can play two games with the remaining teams: 1-2, 1-3, 2-3 (3 fixtures); in division C each team can play 5 games with the remaining teams: 1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6 (15 fixtures); in division D each team can play 7 games: 1-2,...,1-8, 2-3,..., 2-8, 3-4,...,3-8, 4-5,...,4-8, 5-6, 5-7, 5-8, 6-7, 6-8,7-8 (28 fixtures). However, the requirement is 15 games per team. Division C is the only division in which the number of games required for each team to play each of the remaining teams in the same division is a factor of 15, so if the division C fixtures above are played 3 times, each time will have participated in 15 games. Team C1 participates in games 1-5; team C2 participates in games 1, 6-9; team C3 in games 2, 6, 10-12; team C4 in games 3, 7, 10, 13, 14; team C5 in games 4, 8, 11, 13, 15; and team C6 in 5, 9, 12, 14, 15. To participate in 15 games requires 45 games so that each team gets 15 games. In division B, the fixtures would be played 7 times (21 games in all) so that each team gets 14 games; in division D, the fixtures would be played twice (56 games in all), so that each team gets 14 games. What do we about divisions B and D to ensure that each team gets another game to make up to 15?
For divisions B and D we would need to run part of the fixture again so that some teams will exceed the required number of 15 games. In division B we can have 1-2 and 2-3, so that B1 and B3 play 15 games in all, but B2 plays 16 games. That makes 21+2=23 games. In division D we can have 1-2, 3-4, 5-6, 7-8, adding 4 to the 56 to make 60. Each team in division D will then have played 15 games.
Summary: division B plays 23 games; C plays 45 games; and D plays 60. Only one team in division B plays 16 games. 23+45+60=128 time slots are needed.
