At least once is possible - for example:
( 4 10 3) ( 6 9 5) (1 7 2 8)
( 7 6 1) ( 8 3 9) (2 4 5 10)
( 8 2 5) (10 1 9) (7 3 6 4)
(10 8 6) ( 3 1 5) (9 2 7 4)
( 7 10 5) ( 1 8 4) (9 6 3 2)
However it would be nice to do it such a way that all 10 players had exactly two foursomes each. Above players 1,5 & 8 have only one foursome, while players 2,4 & 7 have three foursomes. It may be possible to do slightly better than this.
Actually the problem can be solved in just four rounds. You could take the perfect schedule for 9 players and 4 rounds:
(G F A) (B D C) (E H I)
(F C H) (I G D) (A E B)
(B I F) (D A H) (E C G)
(F D E) (H B G) (C I A)
and then add the 10th player to one of the threesomes in each round such that they play all players A to I at least once. The obvious disadvantage is that one player is always in the foursome.
Hope that helps.
