Hi,
I definitely agree that all-plays-9 is impossible. My experiments on the computer are suggesting that "10 people can play 8 people, 2 can play 7" is also not possible, so I am interested to see the schedule if you have achieved this.
The optimal schedule here is slightly counterintuitive, it has 6 people who play 7 others, and 6 people who play 9 others - it has slightly more unique pairs than you are claiming as 10*8 + 2*7 = 94, while 6*7 + 6*9 = 96. I have given two examples below that you might like to use, where group A are player numbers 1 to 6, and group B are player numbers 7 to 12:
Schedule 1
(1 2 7 8) (3 4 9 10) (5 6 11 12)
(1 3 7 11) (2 5 9 12) (4 6 8 10)
(1 6 7 9) (2 4 10 11) (3 5 8 12)
(1 7) (4 10) & (5 12) always in same foursome
(1 4 5 7 10 12) have 7 different partners
(2 3 6 8 9 11) have 9 different partners
Schedule 2
(1 2 7 8) (3 4 9 10) (5 6 11 12)
(1 2 9 11) (3 4 7 12) (5 6 8 10)
(1 2 10 12) (3 4 8 11) (5 6 7 9)
(1 2) (3 4) & (5 6) always in same foursome
Team A have 7 different partners
Team B have 9 different partners
