Suppose $A = (a_{ij})$ is an $I\times J$ matrix and $a_{ij} \le a_{i'j'}$ if $ i \le i'$ and $j \le j'$. These constraints define a partial order on $\{a_{ij},i=1,\dots,I, j=1,\dots,J\}$ which can be shown by a two dimensional directed lattice. The following picture shows a $3\times 4$ such directed lattice. How many total orderings are there starting from $a_{11}$ to $a_{IJ}$ that includes all the elements of the matrix? For example the following can be a total ordering: $a_{11}, a_{12}, a_{21}, a_{13}, a_{22}, a_{31}, a_{14}, a_{23}, a_{32}, a_{24}, a_{33}, a_{34}$.
Too long for a comment, and I am trying to ascertain if the linked question is a duplicate.
If you take your example ordering, $$ a_{11}, a_{12}, a_{21}, a_{13}, a_{22}, a_{31}, a_{14}, a_{23}, a_{32}, a_{24}, a_{33}, a_{34} $$ and associate each element with its position in the list, so that $a_{11}\gets 1, a_{12}\gets 2,a_{21}\gets 3,a_{22}\gets 4$, etc, and finally place those labels in the lattice, the result is $$ \begin{array}{|c|c|c|c|} \hline 1&2&4&7 \\\hline 3&5&8 &10 \\\hline 6&9&11&12 \\\hline \end{array} $$ This is an example of a standard Young tableau, which in this case is by definition a $3\times 4$ grid filled with the numbers $1$ to $12$ so that the rows increase left to right and the columns increase top to bottom. If we take another standard Young tableau with the same shape, I think we can generate other valid orderings: $$ \begin{array}{|c|c|c|c|} \hline 1&2&3&8 \\\hline 4&6&7 &11 \\\hline 5&9&10&12 \\\hline \end{array}\implies a_{11},a_{12},a_{13},a_{21},a_{31},a_{22},a_{23},a_{14},a_{32},a_{33},a_{24},a_{34} $$ Can you confirm whether or not this last ordering is valid? If so, I claim that all the orderings you want are in bijection with standard Young tableaux of shape $3\times 4$, in which case the other answer works perfectly. If not, can you explain why this does not work?