Multiple rounds of grouping, with no shared groups If I have 20 people, and I want to divide them into 5 groups of 4, can I do that 4 times, such that no one is paired with the same person in a group twice?
E.g. if you labeled each person with a letter, the first round might look like:
ABCD
EFGH
IJKL
MNOP
QRST

And then the next round you could do:
AEIM
FJNR
KOSC
PTDH
BGLQ

such that no one is paired with any of their first-group partners a second time.
But could you repeat this for two more rounds?

If you did a smaller case, say two rounds of three groups of two, it would be easy to construct a valid case. But what sort of formula could you use to check any arbitrary N rounds of M groups of size K ?

I could see one approach would be write quick program to randomly make a million various pairs, and test each one to see if it can find a solution, but can we find a more rigorous way to approach this?
Thanks
 A: It is possible. Let's start by considering the case where you have 25 people instead of 20. Label each of these people with a letter from A-E and a number 1-5.
$A_1, A_2, A_3, A_4, A_5, B_1, ... , C_1, ..., D_1, ..., E_1,...$
In the first round, place all member who have the same letters together (I'll list one example group per round):
$A_1A_2A_3A_4A_5$
In the second round, place all members who have the same numbers together:
$A_1B_1C_1D_1E_1$
In the third round, increment the numbers by 1 for each letter you advance:  
$A_1B_2C_3D_4E_5$
For each subsequent round, you can increase the numbers by 1 more than the previous round, wrapping around if you get greater than 5. For Example:
$A_1B_3C_5D_2E_4$
In this way you can get 6 rounds without 2 members sharing a group twice (This is because 5 is prime).
Now, let's limit it to 4 rounds, since that's what you asked about.
For example: the rounds where groups are the same letter, same number, the difference is 1 and where the difference is 4.
Because we did not use the round where the difference is 2, we can choose one of the groups that would be in that round and eliminate those people. Since they would all be in the same group, they must all be in different groups in previous rounds. Because there are five people in a group and five groups per round, we have eliminated one person from every group.
So, if you eliminate $A_1, B_3, C_5, D_2 $ and $E_4$ from the rounds we selected, we get an example that follows the specified criteria. Based on this, it is possible to actually get a 5th round, since there is another round that we didn't use.
