It’s 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 don’t 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.

Mathematically:

LetSbe the set of all the players in the group. Then a match corresponds to a subset ofScontaining exactly 3 elements (players). Note that it doesn’t 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 ofScontaininga; let be a list of these subsets. In other words, arenmatches in whichaplays.

(i) If you wantato play/partner each of the other 11 players no more than once, each of the other 11 elements ofSmust belong to only one of the subsets . Since 11 is an odd number, it won’t be possible to fit all 11 of them in; there will always be at least 1 element left over. Your best option (i.e. havingaplay/partner as many opponents as possible) is to have and 1 element inSnot in any .

(ii) If you wantato play/partner each of the other 11 players at least once, then all the other 11 elements ofSmust go into somehow. Again it’s because 11 is an odd number that this won’t be possible; this time at least 1 element must belong more than one of the subsets . Your best option this time (i.e. havingadouble-play/partner as few opponents as possible) is to have and 1 element inS(other thana) belonging to and where .