Golf Foursomes: Twenty Players Question: Is it possible to arrange foursomes involving 20 players such that each foursome is unique; i.e., each player is matched with a different player in each of his foursomes, with no repeat players? Answer: Yes. The below shows how to arrange five such rounds.  As in the 16 player case, I will make use of orthogonal Latin squares. (For an explanation of Latin squares, see the 16 player case.) However, because a given player has 19 other players with whom to golf, and 19 is not divisible by three, it is impossible to golf with everyone without incurring a repeat. The below will describe a methodology to match as many players as possible plus a suggestion on what to do with the "remainder". The methodology for twenty players Since we want to form five foursomes for each round, using four 5x5 orthogonal Latin squares would seem to be a logical choice. So, let's try it. First, here are four possible 5x5 Latin squares, mutually orthogonal. Each matrix represents a pool of players to be mixed with the other pools, shown momentarily. Each row represents a different round of golf. Assigning foursomes from the four pools Staring with the first row, and matching the first column of each of the four squares yields the first foursome: AFKP. Repeating this process for columns 2-5 of the first row yields the five foursomes for the first round: Repeating for rows 2-5 yields the following table of foursomes: The rows represent the rounds (five in this case), and the columns represent the five foursomes for each round. Careful inspection should reveal that no two letters, representing players, are repeated. Synopsis With five rounds scheduled above, each player will be partnered with 15 different players in this tournament set-up. That leaves four players that a given player would not have been partnered with during the tournament. In fact, those four players are in the same Latin square as the player of interest. One could form a sixth round by matching players in the same Latin square; however, with five players in a Latin square, one of the players is left out, and to form a fifth foursome, some will be paired with players in a previous foursome. For example, one possibility might be as follows: where the final foursome is a repeat. This leads to 16 players who golf with everyone without incurring a repeat, and four who golf with 15 of the 20 with one foursome being a repeat. Next: Twenty-Four Players