Itís not possible. If there are 12 players, each player will have 11 opponents to play against or partner with, and it is impossible for the player to play/partner every opponent exactly once. If you donít want the player to play/partner any opponent more than once, there must be some opponent that player does not play/partner. If you want the player to play/partner every opponent, there must be some opponent that the player has to play/partner more than once.
Let S be the set of all the players in the group. Then a match corresponds to a subset of S containing exactly 3 elements (players). Note that it doesnít matter who are the partners and who the solo player; all that matters are which three players are involved.
Suppose is a player. There are 3-element subsets of S containing a; let be a list of these subsets. In other words, are n matches in which a plays.
(i) If you want a to play/partner each of the other 11 players no more than once, each of the other 11 elements of S must belong to only one of the subsets . Since 11 is an odd number, it wonít be possible to fit all 11 of them in; there will always be at least 1 element left over. Your best option (i.e. having a play/partner as many opponents as possible) is to have and 1 element in S not in any .
(ii) If you want a to play/partner each of the other 11 players at least once, then all the other 11 elements of S must go into somehow. Again itís because 11 is an odd number that this wonít be possible; this time at least 1 element must belong more than one of the subsets . Your best option this time (i.e. having a double-play/partner as few opponents as possible) is to have and 1 element in S (other than a) belonging to and where .