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I have to organize a meeting with 18 families. Each family has to play 6 plays. For each play 3 families meet.

How can I combine them, so they always meet different families?

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This is not possible; even a simpler variation with 12 families and 2 families per play is impossible. This is known as the Thirty-six officers problem.

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Lets say you had 12 people and 4 rounds and 4 groups. You could accomplish this through the use of orthogonal latin squares!

Here are 3 mutually orthogonal latin squares of order 4.

A B C D    E F G H    I J K L
B A D C    G H E F    L K J I
C D A B    H G F E    J I L K 
D C B A    F E H G    K L I J

Lets superimpose them:

AEI  BFJ  CGK  DHL
BGL  AHK  DEJ  CFI
CHJ  DGI  AFL  BEK
DFK  CEL  BHI  AGJ

You will notice that if these become our groupings, then if you take any one group (say A), and look at its partners, you will never see a duplicate. A plays with E and I in the first round, H and K in the second, F and L in the third, and G and J in the forth.

This satisfies your requirement!!

So all you need to do is find 3 mutually orthogonal latin squares of order 6!

However, this is mathematically impossible.

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