Number of ways in which first $11$ natural numbers can be arranged in the given figure such that $a_{ij}>a_{kj}\;$ whenever $i>k\; \forall \;j$ is ($a_{ij}$ represents elements in $i^{th}$ row and $j^{th}$ column)
Attached is image for the same
My thinking till Now:
Assuming $a_{41}>a_{45}\;$, $\;a_{31}>a_{33}>a_{34}\;$, $a_{21}>a_{22}>a_{24}>a_{25}\;$, and $a_{11}>a_{15}$
I obtained $a_{41}>a_{45}>a_{31}>a_{33}>a_{35}>a_{21}>a_{22}>a_{24}>a_{25}>a_{11}>a_{15}\;$.
Now $a_{41}>a_{45}\;$, $\;a_{31}>a_{33}>a_{34}\;$, $a_{21}>a_{22}>a_{24}>a_{25}\;$, and $a_{11}>a_{15}$ can be arranged in $2!\cdot 3! \cdot 4! \cdot 2!$
So total ways i obtained is $576$ but answer given is $69300$
The only constraint is that values increase down the columns. So choose $4$ values for the first column and then $4$ values for the last column. There is then only one way to arrange each of these columns. The remaining three numbers can be placed arbitrarily in the middle three squares.
So the total number of arrangements is given by: $${{11}\choose{4}}\cdot {{7}\choose{4}}\cdot 3! $$
This seems to be consistent with the given answer of $69300$.