Let's say there is a golf tournament going on with $6$ players. Everyone plays $4$ rounds and they play with each other in groups of $3$ each time. Everyone has to play with each other either once or twice (but no other amount). Is this possible? And if so then how?
I've tried to solve this using combinations and I've also tried to program it. Both to no avail. I have found a solution where everyone plays each other but some play with each other $3$ times (which isn't allowed). So please can someone help me with this? Thanks.
It is indeed impossible.
Without loss of generalization, we can assume that round $ 1 $ is:
$1,2,3$
$4,5,6$
If for round $2$ we have the exact same groups, we immediately have a problem, because then for round $3$ player $ 1$ needs to play with two players from $4,5,6$ who already played twice with each other.
So, in round $2$ either one or two players from one group goes to the other group, but since two players switching groups amounts to the same thing as one player (the third) switching groups, we can again without loss of generalization say that round $2$ has to be:
$1,2,4$
$3,5,6$
OK, so now $1$ and $2$ have played each other twice, and same for $5$ and $6$, so both these pairs need to be split up.
Moreover, $ 1 $ still needs to play with $ 5 $ and with $ 6$, so one round will be with $5$ and the other one with $6$. So the last two rounds have to look like (order does not matter):
$1,5,X$ ($X$ is $ 3 $ or $4$)
$2,6,Y$ ($Y$ is $3$ or $4$)
and:
$1,6,V$ ($V$ is $3$ or $4$)
$2,5,W$ ($W$ is $3$ or $4$)
Well, that's impossible: since $1$ already played one time each with $3$ and $4$, $X$ and $V$ have to be different, and thus $Y$ and $V$ are the same. But $6$ has played one time each with $3$ and $4$ as well, meaning that $6$ ends up playing with one of them three times.