Number of combinations to introduce people within certain groups

Question

I am trying to calculate a combination of unique people inside of the different groups. Nonetheless, I have not observed any formula to achieve my intention. There are 4 different groups and each group has 4 people. Besides, the main question here is, what is the total number of combinations that I can introduce to every people by picking one person from each group to a new groups of 4 people? I want every people to meet each other and the total number of enumerations are needed to compute the total solution.

For instance;

Group 1 = $\mathit {A,B,C,D}$

Group 2 = $\mathit {H,I,J,K} $

Group 3 = $\mathit{M,N,O,P}$

Group 4 = $\mathit{X,Y,Z,T}$

NewGroup1 = $\mathit{A,H,M,X}$

NewGroup2 = $\mathit{B,I,N,Y}$

NewGroup3 = $\mathit{A,I,O,Z}$

...

How many NewGroups will be?

Note: I was thinking about applying a minimum colouring problem but this might turn out to be a NP-Hard problem. I also have in mind that to use permutation with repetition, but I am not sure it will give the correct result.

Answer 1

What you are looking for is a special case of a covering design; a collection of subsets of size $k=4$, from a set of size $v=16$, such that every group of size $t=2$ appears together in some subset. Looking at the $(v,k,t)=(16,4,2)$ entry of the La Jolla Covering repository, there is a solution using only $20$ groups, including the four initial groups. Here is the result, after some renaming and reorganizing to fit your problem.

Your particular problem can actually be solved by three orthogonal Latin squares of order four.

A, B, C, D, 
H, I, J, K,  
M, N, O, P, 
X, Y, Z, T, 

X, C, K, M, 
H, C, T, P, 
H, B, Z, O, 
H, A, X, N, 
H, D, M, Y, 
A, I, M, Z,
K, D, Z, P, 
B, K, Y, N, 
A, J, Y, P, 
J, C, Z, N, 
B, I, X, P, 
J, X, D, O, 
A, K, O, T, 
I, D, N, T,
I, C, Y, O, 
B, J, M, T,