The following problem is open in the sense that no solution is currently known. An explicit expression with proof for the known case of $|A(5, n)|$ will be awarded 2 points. Further progress on the problem may be awarded more points.

For positive integers $m$ and $n$, let $A(m, n)$ be the set of $2 \times m n$ matrices $M$ with entries from the set $\{1,2, \ldots, m\}$ such that
- each of the numbers $1,2, \ldots, m$ appears exactly $2 n$ times;
- $M_{1,1} \leq M_{1,2} \leq \cdots \leq M_{1, m n}$ and $M_{2,1} \leq M_{2,2} \leq \cdots \leq M_{2, m n}$; and
- $M_{1, j}<M_{2, j}$ for $j=1,2, \ldots, m n$.

Determine $|A(m, n)|$.
Solution to the case $m=5$
Let $M$ be a matrix satisfying the given conditions with $m=5$ and $n$ a positive integer. Then we must have $M_{1, i}=1$ for $1 \leq i \leq 2 n$ and $M_{2, j}=5$ for $3 n+1 \leq j \leq 5 n$. Note that each matrix is uniquely determined by its first row, since the second row must consist of the remaining numbers in non-decreasing order.

Let the number of $2 \mathrm{~s}, 3 \mathrm{~s}$ and $4 \mathrm{~s}$ in the first row be $x_{2}, x_{3}$ and $x_{4}$, respectively. So the number of $2 \mathrm{~s}, 3 \mathrm{~s}$ and $4 \mathrm{~s}$ in the second row must be $2 n-x_{2}, 2 n-x_{3}$ and $2 n-x_{4}$, respectively. Therefore, we have the inequality
$$
0 \leq x_{2}, x_{3}, x_{4} \leq 2 n
$$

Since 1 appears $2 n$ times in the first row and 5 does not appear in the first row, we must have
$$
x_{2}+x_{3}+x_{4}=5 n-2 n=3 n \text {. }
$$

Furthermore, the number of $2 \mathrm{~s}$ and $3 \mathrm{~s}$ in the second row must not exceed the number of $1 \mathrm{~s}$ and 2 s in the first row. So we obtain the inequality
$$
2 n+x_{2} \geq\left(2 n-x_{2}\right)+\left(2 n-x_{3}\right) \quad \Rightarrow \quad 2 x_{2}+x_{3} \geq 2 n
$$

One can show that every integer solution to the system
$$
0 \leq x_{2}, x_{3}, x_{4} \leq 2 n, \quad x_{2}+x_{3}+x_{4}=3 n, \quad 2 x_{2}+x_{3} \geq 2 n,
$$
leads to a a matrix satisfying the given conditions and vice versa. So $|A(5, n)|$ is precisely equal to the number of integer solutions to the system.

If we fix $x_{2}=k$ for $0 \leq k \leq 2 n$, then the system becomes
$$
0 \leq x_{3}, x_{4} \leq 2 n, \quad x_{3}+x_{4}=3 n-k, \quad x_{3} \geq 2 n-2 k .
$$

Using $0 \leq x_{4} \leq 2 n$ in the equation $x_{3}+x_{4}=3 n-k$ leads to $n-k \leq x_{3} \leq 3 n-k$. Therefore, we obtain
$$
\max (0, n-k, 2 n-2 k) \leq x_{3} \leq \min (2 n, 3 n-k),
$$
and each such $x_{3}$ leads to a unique value for $x_{4}$.
We now complete the calculation of $|A(5, n)|$ as follows.
$$
\begin{aligned}
|A(5, n)| & =\sum_{k=0}^{2 n}[\min (2 n, 3 n-k)-\max (0, n-k, 2 n-2 k)+1] \\
& =\sum_{k=0}^{n}[(2 n)-(2 n-2 k)+1]+\sum_{k=n+1}^{2 n}[(3 n-k)-(0)+1] \\
& =\sum_{k=0}^{n}(2 k+1)+\sum_{k=n+1}^{2 n}(3 n-k+1) \\
& =\frac{(n+1)(2 n+2)}{2}+\frac{n(3 n+1)}{2} \\
& =\frac{1}{2}\left(5 n^{2}+5 n+2\right)
\end{aligned}
$$

Comments
This combinatorics problem arises naturally in the study of certain spaces that arise in algebraic geometry - namely, quotients of Grassmannians by torus actions. For some related combinatorics problems, please see sequence A060854 at the The On-Line Encyclopedia of Integer Sequences (https://oeis.org/A060854).