Counting sets which can be partition in two subsets satisfying certain criterion I have a question that involves some ability in combinatorics that I don't have. Any hint could be extremely appreciated.
Take $n>1$ with $n\in \mathbb{N}$.
Let $\mathcal{G}$ be the set of all possible binary matrices of dimension $n\times (n-1)$. $\mathcal{G}$ has cardinality $|\mathcal{G}|=2^{n(n-1)}$. 
For example, if $n=2$, then $\mathcal{G}:=\{\begin{pmatrix}
1\\
1
\end{pmatrix},\begin{pmatrix}
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1
\end{pmatrix} , \begin{pmatrix}
0\\
0
\end{pmatrix}\}$ with $|\mathcal{G}|=4$. 
Question: For $M=2,...,|\mathcal{G}|-1$, I want to count all possible sets $C\subset \mathcal{G}$ with cardinality $M$ such that:
there exists a non-empty set $D\subset C$ such that for every pair of matrices, one taken from $D$ and the other taken from $C-D$, the two matrices in the pair considered differ for at least two rows.
Is this counting possible? If not, is there any known upper bound or lower bound?
I have looked at this question whose comments link to bounds for binary constant weight codes, but it does not seem to help.
[$C-D$ denotes the complement of $D$ in $C$].
For example: when $n=2$ the result is $2$. Indeed, for $M=2$, we can have
$$
C=\{\begin{pmatrix}
1\\
1
\end{pmatrix}, \begin{pmatrix}
0\\
0
\end{pmatrix}\} 
$$
where $D=\{\overbrace{\begin{pmatrix}
1\\
1
\end{pmatrix}}^{d_1}\}$, $C-D=\{\overbrace{\begin{pmatrix}
0\\
0
\end{pmatrix}}^{c_1}\}$, row $1$ of $d_1$ $\neq $ row $1$ of $c_1$, and row $2$ of $d_1$ $\neq $ row $2$ of $c_1$
and
$$
C=\{\begin{pmatrix}
0\\
1
\end{pmatrix}, \begin{pmatrix}
1\\
0
\end{pmatrix}\} 
$$
where $D=\{\overbrace{\begin{pmatrix}
0\\
1
\end{pmatrix}}^{d_1}\}$, $C-D=\{\overbrace{\begin{pmatrix}
1\\
0
\end{pmatrix}}^{c_1}\}$, row $1$ of $d_1$ $\neq $ row $1$ of $c_1$, and row $2$ of $d_1$ $\neq $ row $2$ of $c_1$
For $M=3$, there is no $C$ satisfying the considered criterion.
Remark: Notice that the order of the rows matter. 

This is not an answer but an input regarding the maximum of $M$.
Claim:
For $M \gt 2^{n(n-1)}-n(2^{n-1}-1)$ there is no $C$ satisfying the criterion. 
Argument:
When pairing matrices, at least two rows must differ. This means the only pairing which isn't allowed is the pairing of matrices which differ in exactly one row. For a given matrix, how many such disallowed pair-mates are there? 
If a row has $n-1$ binary digits there are $2^{n-1}$ possible permutations of that row. For a given row and a given matrix, there are thus $2^{n-1}-1$ matrices which differ from the given matrix in only that row. With $n$ rows there are thus $n(2^{n-1}-1)$ matrices which differ from the given matrix in one row. 
An example with $n=3$. The given matrix:
$$
        \begin{pmatrix}
        0 & 0 \\
        0 & 0 \\
        0 & 0 \\
        \end{pmatrix}
$$
has the following $n(2^{n-1}-1)$ disallowed pair-mates:
$$
        \begin{pmatrix}
        0 & 1 \\
        0 & 0 \\
        0 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        1 & 0 \\
        0 & 0 \\
        0 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        1 & 1 \\
        0 & 0 \\
        0 & 0 \\
        \end{pmatrix}       
$$
$$
        \begin{pmatrix}
        0 & 0 \\
        0 & 1 \\
        0 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        0 & 0 \\
        1 & 0 \\
        0 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        0 & 0 \\
        1 & 1 \\
        0 & 0 \\
        \end{pmatrix}       
$$
$$
        \begin{pmatrix}
        0 & 0 \\
        0 & 0 \\
        0 & 1 \\
        \end{pmatrix}\begin{pmatrix}
        0 & 0 \\
        0 & 0 \\
        1 & 0 \\
        \end{pmatrix}\begin{pmatrix}
        0 & 0 \\
        0 & 0 \\
        1 & 1 \\
        \end{pmatrix}       
$$
This means a given matrix can only be paired with $2^{n(n-1)}-n(2^{n-1}-1)-1$ other matrices (the $-1$ at the end is because a given matrix cannot be paired with itself). A subset $C$, where $D$ consists only of the given matrix, can therefore at most have $|C-D| = 2^{n(n-1)}-n(2^{n-1}-1)-1$ other elements and can therefore at most have cardinality $M = 2^{n(n-1)}-n(2^{n-1}-1)$.
If the sole element of $D$ is the given matrix, we know from above that we cannot add extra matrices in $C-D$, but perhaps we can add extra matrices in $D$, thus increasing $M$? 
If we add a disallowed matrix to $D$ (say the upper left matrix in the example above) we see that this new member of $D$ doesn't share $(n-1)(2^{n-1}-1)$ of the disallowed with the given matrix. Since the total number of matrices that a matrix can be paired with remains constant for a given $n$, the number of matrices that both the given matrix and the new member can be paired with, is reduced by $(n-1)(2^{n-1}-1)$. So one extra member of $D$, but a greater reduction in $|C-D|$.
If we add an allowed matrix to $D$, the reduction of $|C-D|$ would be even greater.
Let me know what you think.
