How many rounds would it take to get each pair on the same team at least once, not using all possible teams? I have a young group of kids ($30$) playing soccer and they need to be put into $6$ teams of $5$ players for each round of matches. All $6$ teams play at the same time on adjoining fields.
If I wanted each kid to play on a team with every other kid in the group (so they know each others names), how many team rounds would I need?
Note: I'm not after the number of rounds to get through all of the combinations of unique teams.
 A: Here's a $10$-round solution with some theory behind it, although it seems like the theory hasn't helped us do better than a computer search.

Let's divide the $30$ kids into four groups: group abcdefghijkl, group 123456, group OPQRST, and group UVWXYZ. Then, begin with the following six rounds:
ab1PX   cd2QY   ef3RZ   gh4SU   ij5TV   kl6OW
ab2RU   cd3SV   ef4TW   gh5OX   ij6PY   kl1QZ
ab6SW   cd1TX   ef2OY   gh3PZ   ij4QU   kl5RV
ab3TY   cd4OZ   ef5PU   gh6QV   ij1RW   kl2SX
ab5QZ   cd6RU   ef1SV   gh2TW   ij3OX   kl4PY
ab4OV   cd5PW   ef6QX   gh1RY   ij2SZ   kl3TU

Where do these come from? I have built this out of the three mutually nearly orthogonal Latin squares of order $6$ constructed on p.24 of this paper, with one Latin square used on the group 123456, one used on OPQRST, and one on UVWXYZ.
If we took actually orthogonal Latin squares, then the result would be that all kids in different groups have met. Unfortunately, orthogonal Latin squares of order $6$ don't exist. This set has almost that property: the six triples 1OU, 2PV, 3QW, 4RX, 5SY, 6TZ are all we need.
In four more rounds, we can finish the job, and also have the three small groups meet each other; this can be done by hand easily without too much thought. Here is the partial construction of these four rounds:
2PV..   3QW..   4RX..   5SY..   .....   .....
12345   UZ6T.   OPQRS   .....   .....   .....
23456   UVWXY   OT...   .....   .....   .....
16OU    VWXYZ   PQRST   .....   .....   .....

Having everyone in the large group meet (that hasn't already) is harder; this is the bottleneck, and can't be done in $3$ rounds. For this, I gave in and did a simulated annealing run of my own, and got the following four rounds (I represent with . positions that can be filled in any way you like):
2PV..   3QW..   4RX..   5SY..   aegil   bfhjk
12345   UZ6T.   OPQRS   acfgk   bdhil   ej...
23456   UVWXY   OT...   achjl   defik   bg...
16OU.   VWXYZ   PQRST   adgjk   bcehi   fl...

These four rounds, together with the six rounds at the beginning, are a $10$-round solution to the problem.

I can't prove that there's no $9$-round solution. The Latin-square based approach can't possibly be made to work in $9$ rounds, however, and it seems very efficient (except that the six pairs ab, cd, ef, gh, ij, kl are stuck together all the time).
