TITLE: How many rounds would it take to get each pair on the same team at least once, not using all possible teams?
QUESTION [21 upvotes]: 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.

REPLY [5 votes]: This answer is just a supplement to the answer by Yorch, but it's a bit too large to fit into a comment. ;)
The 8 round partial solution has pair coverage (0: 14, 1: 365, 2: 53, 3: 3). That is, only 14 pairs are uncovered, 365 pairs are covered once, etc. It has perfect triplet coverage: 480 triplets are covered exactly once. (Each round covers 60 triplets).
I completed the 9th round by simply assigning unused numbers to the partial and empty groups. This produced a final row of
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 7, 8, 21, 23], [5, 9, 14, 19, 22], [11, 13, 15, 16, 18], [20, 24, 25, 26, 27]] 

which has coverage (1: 341, 2: 83, 3: 11)
By randomly shuffling the unused numbers, we can increase the number of pairs that have coverage of 1, and reduce the pairs with coverage 2, but that increases the number of pairs that are covered 3 times.
Here are typical results, found by searching through 10000 random permutations. I actually searched 500000 permutations, but I didn't find any solutions which have >346 pairs with coverage 1.
8 classes
0: 14, 1: 365, 2: 53, 3: 3, 

[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 7, 8, 21, 23], [5, 9, 14, 19, 22], [11, 13, 15, 16, 18], [20, 24, 25, 26, 27]]
1: 341, 2: 83, 3: 11, 
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 16, 21, 23, 26], [5, 14, 19, 22, 24], [8, 9, 13, 15, 18], [7, 11, 20, 25, 27]]
1: 342, 2: 81, 3: 12, 
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 8, 21, 23, 25], [5, 7, 14, 19, 22], [11, 15, 16, 26, 27], [9, 13, 18, 20, 24]]
1: 343, 2: 79, 3: 13, 
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 7, 16, 21, 23], [5, 14, 19, 22, 26], [9, 13, 18, 20, 24], [8, 11, 15, 25, 27]]
1: 344, 2: 77, 3: 14, 
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 8, 16, 21, 23], [5, 14, 19, 22, 24], [7, 11, 15, 18, 26], [9, 13, 20, 25, 27]]
1: 345, 2: 75, 3: 15, 
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 7, 16, 21, 23], [5, 14, 19, 22, 24], [8, 13, 20, 25, 27], [9, 11, 15, 18, 26]]
1: 346, 2: 73, 3: 16, 

Going in the other direction, we reduce the number of pairs that are covered once or 3 times, but increase those with a coverage of 2. Eg,
[[0, 2, 10, 17, 29], [1, 3, 6, 12, 28], [4, 9, 21, 23, 24], [5, 14, 19, 22, 25], [7, 15, 16, 18, 20], [8, 11, 13, 26, 27]] 

which has coverage (1: 336, 2: 93, 3: 6).
You can see and run my Python code here.
FWIW, a 9 round equitable covering would have coverage (1: 330, 2: 105), and an 8 round equitable covering would have coverage (1: 390, 2: 45).