Say we have a group containing 4 items $[A, B, C, D]$ and we allow sub-groups of size 2, then 6 sub-groups are needed to ensure every element is grouped with every other element at least once.
$[A,B] [A,C] [A,D] [B,C] [B,D] [C,D]$
or 3 if we allow sub-groups size 3
$[A,B,C] [A,B,D] [B,C,D]$
Expanding this to 8 elements and sub-groups size 4, 6 sub-groups are also required.
It also appears that if you have $n$ elements and a sub-group size $m$ which require $x$ sub-groups to ensure element grouping, then $kn$ elements with sub-group size $km$ also requires $x$ sub-groups. So possibly only prime numbers for $n$ or $m$ have unique values of $x$.
I tried initially looking at binary representations as a way to get all possible arrangements, then counting the instances of Hamming Weight $m$ from $0$ to $2^n$ but the pattern did not hold.
I am trying to generalize it so if you have $n$ elements with sub-group size $m$
$\binom{n}{m}$ where $1<m<n$ , $m \in \mathbb{Z}$ , $n \in \mathbb{Z}$
some function $F(n,m) = x$
Is there a formular to express the minimum number of sub-groups size $m$ so that each element in $n$ is grouped with every other element at least once?
I'm having trouble describing elements as having already been grouped mathematically.