Mathematical dilemma

[jeraldo]

There are 10 people who want to play golf over 4 games so that each one plays with everyone else at some time. There will be 2 groups of 3 and one group of 4 each day. Is this possible? I have been trying to work it out without any form of mathematical logic. Can anyone help?

[factor-fiction]

I don't think it can be done. I'll explain my reasoning shortly- I just need to check it out on paper.

[factor-fiction]

Call the 10 people A, B, C, D, E, F, G, H, I, J

With 10 players, at any one time there will be 2 groups of 3 and one group of 4

To play everyone once each player has to play against 9 others and a player can never play with the same person twice.

So for example A has to play with B &C, then with D &E, then with F&G then with H,I &J. Each player needs to be in 3 groups of 3 (himself plus 2) and one group of 4 (himself plus 3)

Suppose on day 1 the groups are:
A B C, DEF and GHIJ

On day 2: A has to find 2 new partners and goes with E. They need a third player. Because A has already played with B & C , and D has already played with E&F, then they have to choose one of GHI&J. Let's suppose it is G. So the group is AEG

Also on day 2 B chooses 2 partners. He chooses D (from day 1 group DEF) and H. So the group is BDH.

That leaves CFIJ as the foursome. But I and J have already played together. You can't add I or J to the other groups either as they have already played G and H.

[jeraldo]

Thanks factor 30. Actually, they can play together twice, which makes it that much easier to caluculate ... does this help?

[jeraldo]

... and the other element was that I wanted everyone to play both 3 and 4 ball games - again, it doesn't matter if they do 3 of one and one of another but two and two would be good ...

[factor-fiction]

No - if anyone plays someone twice it means they can't get round to playing with everyone.

Each day you have 3 groups- 2 groups of 3 and one group of 4. Agreed?

Take yourself. You have to play with 9 different people over 4 days. That means you have to play in 3 threesomes (excluding yourself that means a total of 6 different players) and one foursome (3 different players). That's the only way to play 9 people over 4 days using these group sizes. There's no time to play with anyone twice- unless you miss playing with someone else.
Agreed?

[jeraldo]

Yes - agreed. The main criterion was that everyone played with everyone at sometime and I reckoned, like you, that if one player played 2 x 3 balls and 2 x 4 balls, they would play with everyone at least once - didn't matter if they played with one person two or even three times! But I couldn't come up with a satisfactory structure.

The potential saving grace (which I don't want to take into account) is a final 5th round but that is normally based on an order of merit from the previous rounds - ie last place out first and so on until the best placed goes last and theoretically the top players finish last in grandstand order! It may be if things fell right there would be some luck on the fifth round but I really wanted to confine things to the first four if at all possible!

Does this make sense