Formula for combinations

[NeedmathhelpinSeattle]

Question: I have a group of 12 people. I would like to break them into 3 groups of 4, for a total of 4 rounds (each with 3 groups of 4). I would like for everyone to have an opportunity to work with each of the other 11 people at least once. What equation can i use to break up the groups? Thank you.

 

[Steven G]

I think that you have the wrong idea of this forum. The helpers here are the ones who ask the posters leading questions so that they can solve their own problem. To receive help you need to post a clearly written problem, show your work and explain why you are having trouble.

 

[Cubist]

I'm pretty sure that OP isn't a student. @NeedmathhelpinSeattle please tell us a little bit more about your background in mathematics (and if you are a student then what level are you studying at). Where did you come across this problem?

 

[pka]

Question: I have a group of 12 people. I would like to break them into 3 groups of 4, for a total of 4 rounds (each with 3 groups of 4). I would like for everyone to have an opportunity to work with each of the other 11 people at least once. What equation can i use to break up the groups? Thank you.

In addition to telling us about yourself, please tell us what you know about ordered and/or unordered partitions.

 

[Cubist]

Question: I have a group of 12 people. I would like to break them into 3 groups of 4, for a total of 4 rounds (each with 3 groups of 4). I would like for everyone to have an opportunity to work with each of the other 11 people at least once. What equation can i use to break up the groups? Thank you.

It's a shame that you've chosen to remain silent, since responding to our questions would've speeded things up . Here's the best result that I found (using a computer to search for the fewest zeros in the table below) with 4 rounds...

Round details
=============
 0 1 2 3 : 4 5 6 7 : 8 9 A B
 0 1 4 5 : 2 6 8 9 : 3 7 A B
 0 1 6 A : 2 4 7 8 : 3 5 9 B
 0 1 7 9 : 2 5 8 A : 3 4 6 B


Counting the times that a player is with another player
=======================================================

      1 2 3 4 5 6 7 8 9 A B
    -----------------------
 0:   4 1 1 1 1 1 1 0 1 1 0
 1:     1 1 1 1 1 1 0 1 1 0
 2:       1 1 1 1 1 3 1 1 0
 3:         1 1 1 1 0 1 1 3
 4:           2 2 2 1 0 0 1
 5:             1 1 1 1 1 1
 6:               1 1 1 1 1
 7:                 1 1 1 1
 8:                   2 2 1
 9:                     1 2
 A:                       2
 B:

Histogram of the above counts
=============================
 0:8  1:48  2:7  3:2  4:1

I don't gaurantee these results are correct (or the best) since there might be a bug in the code that I wrote ?. Intuitively, it seems like it would be possible to do better. However, I won't be doing any extra research into this unless you respond.