scheduling a tournament with constraints I want to schedule a wrestling tournament with $7$ teams, where the teams meet once a week for $4$ weeks,
and each week there are $4$ teams at school $M$ and $3$ teams at school $N$.
I would like to find out if it is possible to schedule the teams so that 
1) every pair of teams is at the same school for at least one week and
2) each team is at school $M$ at least twice.
I haven't been able to produce a schedule satisfying both conditions, so is this  possible?
 A: There are 4 meetings with partition $4+3$ so that we will have at most $$4\cdot \left(\frac{4\cdot 3}{2} + \frac{3 \cdot 2}{2}\right) = 36$$ pairs possible. On the other hand, we have to accommodate $\frac{7\cdot 6}{2} = 21$ pairs, meaning, that we can have at most $36-21 = 15$ instances of previously used pairs.
On the other hand, we have to satisfy requirement "each team at school $M$ at least twice" and that "each pair meets at least once". There are $16$ slots for $M$ and $14$ double visits, at least 5 teams have to visit $M$ exactly twice.
Observe, that for each pair of teams which visit $M$ exactly twice, they have to meet at least twice, once in $M$ and once in $N$. That means, every such pair produces one redundant pair as well, thus 10 redundant pairs.
Now we have two cases, either there is a team $G$ which never visits $N$, or there are two teams $F,G$ who visit $M$ three times. In the first case that team it produces $6$ additional redundant pairs (one with every other team), for a total of at least $16$, so that's impossible.
In the second case if teams $F,G$ doesn't visit $M$ on the same days (thus produce only 1 redundant pair), we have only 1 non-redundant-pair schedule (the one that visits $M$ when $F$ and $G$ are non visiting $M$). Any other schedule gives two additional redundant pairs with at least one of $F$ or $G$, and repeated non-redundant-pair schedule also gives two additional redundant pairs.
In total that is at least $10+1+2\cdot 4 = 19$, contradiction.
Yet, if $F$ and $G$ visit $M$ on the same days (3 more redundant pairs), then there are only 3 non-redundant-pair schedules (for 5 teams) – either two schedules are the same (2 additional redundant pairs each) or some teams use other schedules (2 additional redundant pairs each), for a total of at least $10+3+2\cdot 2 = 17$, also impossible.
Therefore, all the cases imply we would need at least $16$ redundant pairs, which means there is not enough time/space for non-redundant pairs, and there is no assignment that satisfies given conditions.
I hope this helps $\ddot\smile$
