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Question from Steven:

We have 24 golfers (6 A's, 6 B's, 6 C's & 6 D's) playing 4 days.
Each day a foursome consist of an A, a B, a C & a D player.
Can you give me a pairing list so that no two golfers will play on the same foursome for the 4 days?

Steven, we have two responses for you

Write 4 rows of six numbers:

A: 1, 2, 3, 4, 5, 6
B: 1, 2, 3, 4, 5, 6
C: 1, 2, 3, 4, 5, 6
D: 1, 2, 3, 4, 5, 6

On the first day all the 1's play together, as do all the 2's, etc.
On the second day its A1, B2, C3, D4; A2, B3, C4, D5; A3, B4, C5, D6; A4, B5, C6, D1; A5, B6, C1, D2; A6, B1, C2, D3
On the third day do something similar to day to but "cycle left" starting with A1, B6, C5, D4; A2, B1, C6, D5, etc

Because 6 is a number with many divisors, it won't work out to use a similar pattern for day 4. But it isn't hard to find something that works by playing around a bit.

Good luck.


Look at the response I gave a while ago to a "24 golfers schedule".

Also there is a web page called the Social Golfer Problem put up by Wolfram Research that allows you to solve many golf scheduling problems.


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