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 Question from Duane, a parent: Got 24 golfers playing golf for 9 rounds. Any formula where everyone can play with everyone else at least once. We are playing 4somes only. Thanks, Duane

Duane,

Try this. Break the golfers into four groups of six. Say:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

The first collection of foursomes is the columns: 1, 7, 13, 19; 2, 8, 10 , 14; etc.
The second set of foursomes is 1, 8, 15, 22; 2, 9, 16, 23; 3, 10, 17, 24; 4, 11, 18, 19; 5, 12, 13, 20; 6, 7, 14, 21
The third set is 1, 9, 17, 19; 2, 10, 18, 20; 3, 11, 13, 21; 4, 12, 14, 22,; 5, 7, 14, 23; 6, 8, 16, 24
The fourth set is 1, 10, 13, 22; 2, 11, 14, 23; 3, 12, 15, 24; 4, 7, 16, 19; 5, 8, 17, 20; 6, 9, 18, 14
The fifth set is 1, 11, 15, 19; 2, 12, 16, 20; 3, 7, 17, 21; 4, 8, 18, 22; 5, 9, 13, 23; 6 10, 14, 24
The sixth set is 1, 12, 17, 22; 2, 7, 18, 23; 3, 8, 13, 24; 4, 9, 14, 19; 5, 10, 15, 20; 6, 11, 16, 21

This takes case of all pairs belonging to different groups of six. Now for the rest. I think this works:
Day 7: 1, 2, 3, 4; 5, 6, 7, 8; 9, 10, 11, 12; 13, 14, 15, 16; 17, 18, 19, 20; 21, 22, 23, 24
Day 8: 3, 4, 5, 6; 7, 8, 9, 10; 11, 12, 13, 14; 15, 16, 17, 18; 19, 20, 21, 22; 23, 24, 1, 2
Day 9: 5, 6, 1, 2; 7, 8, 11, 12; 13, 14, 17, 18; 19, 20, 23, 24; 3, 4, 9, 10; 15, 16, 21, 22

In the last three days some pairs (ex. 1 and 2) are together every day. There are 12 such pairs. There might be a better way to do it. Can you see one? Otherwise, keep friendships in mind when numbering the players.

Have fun.
Victoria

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