I have 8 people playing 4 rounds of golf. Playing each round in two groups of four people is it possible that all players play together twice? Trying to work out a draw for it.
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1In a $t-(v,k,\lambda)$ block design we always have $\lambda\binom{v}{t}=b\binom{k}{t}$. In this case the lhs is $2*\binom{8}{2}=56$ while the rhs is $8\binom{4}{2}=48$. So no such design exists. – Asinomás Feb 24 '15 at 05:49
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No. Suppose you have a person $A$ with $7$ friends. Over the course of $4$ rounds of golf, $A$ will have played in a group with exactly $12$ other people. But since he has $7$ friends, he would have to play with $14$ people to play with each twice.
Mike Pierce
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Is there a way to get it close? At the moment some people play together 3 times and zero with others. – MARK Feb 24 '15 at 05:49