Golf Foursomes: Twenty-Four Players Question: Is it possible to arrange foursomes involving 24 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 and 20 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 23 other players with whom to golf, and 23 is not divisible by three, it is impossible to golf with everyone without incurring a repeat. So, as in the 20 player case, the below will describe a methodology to match as many players as possible before incurring a repeat. The methodology for twenty-four players In the 20 player case, I made use of four 5x5 matrices, three of which were mutually orthogonal Latin squares and the fourth being orthogonal to the other three, though not a true Latin square. So, to extend the methodology from 20 to 24, one might be attempted to graduate to 6x6 orthogonal Latin squares. However, 6x6 Latin squares do not exist. This was first postulated by Leonard Euler, the famous 19th century mathematician. In fact, he was so sure that no Latin squares existed for orders 2, 6, 10, 14... and indeed for all orders of the form 4n+2, that this was given a name: Euler's Conjecture. In 1900, G. Tarry proved that no orthogonal squares of order 6 exist thus lending credibility to the Euler's Conjecture 1. (In 1960, Bose, Shrikhande, and Parker showed that, except for this one case, the conjecture was false 2.) So, since we cannot use an approach based on order 6, what are we to do? Actually, what we can do is base a solution upon the approach in the 20 player case: namely, make use of the order 5 approach, and make an extension to it. First, here are 4 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. Players A-T covers 20 of our 24 players. To include our remaining 4 players, I'll list them as a "remainder" foursome at the end of each round. Now, taking advantage of the fact that players in the same pool will never be paired together, I'll rotate the remainder into each pool, one round at a time (starting with the second round), as follows: This establishes our pools of players from which we will assign foursomes, described next. Assigning foursomes from the four pools The process for assigning foursomes is straight-forward as in the previous cases. Staring with the first row, and matching the first column of Pools 1-4 yields the first foursome: AFKP. Repeating this process for columns 2-5 of the first row of Pools 1-4 yields the five foursomes for the first round, including the remainder foursome: 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) appear twice in a given foursome. Synopsis The above methodology results in a given player being matched with 15 other players without incurring a repeat. All subsequent rounds will incur a repeat pairing of one player with another somewhere. The problem is to decide how to spread out the repeats and and the same time add players not previously matched. Trial and error is one way to go. __________________________________ 1 Tarry, G. "Le problème de 36 officiers." Compte Rendu de l'Assoc. Français Avanc. Sci. Naturel 1, 122-123, 1900. 2 Bose, R. C.; Shrikhande, S. S.; and Parker, E. T. "Further Results on the Construction of Mutually Orthogonal Latin Squares and the Falsity of Euler's Conjecture." Canad. J. Math. 12 , 189, 1960.