Essentially there are no nice solutions to this problem. If I concentrate on the group of 7 players who are supposed to play each other, then I believe the only way for them to do this, within the format that you have outlined, is as follows:
R1 (- - 1) (2 3 4) (- 5 6 7)
R2 (- 3 5) (- 4 6) (- 1 2 7)
R3 (- 4 7) (- 2 5) (- 1 3 6)
R4 (- 2 6) (- 3 7) (- 1 4 5)
where dashes represent slots for the other 3 players. The problem now is that there is no nice way to place the remaining 3 players in the empty slots, you will quickly find that you can not avoid having some pairs play together 3 times. Possibly the best compromise here is to play:
R1 (A B 1) (2 3 4) (C 5 6 7)
R2 (A 3 5) (B 4 6) (C 1 2 7)
R3 (A 4 7) (B 2 5) (C 1 3 6)
R4 (A 2 6) (B 3 7) (C 1 4 5)
where A, B, & C are the remaining players such that (C 1) is the only pair that is repeated 3 times. But I can see you might not like the fact that 'C' always plays in the foursome. Hope that helps.
