The number of ways in which a $9$-by-$9$ grid can be filled given some conditions. 
How to find the number of ways in which the above $9$-by-$9$ grid can be filled using the digits $(1-9)$ (repetition is allowed) such that all of the following conditions are satisfied:


*

*Any $3$-by-$3$ grid is not totally empty.

*Any $3$-by-$3$ grid is not totally filled.                 

*Any $3$-by-$3$ grid must not contain repeated digits.

*Any row must not contain repeated digits.

*Any column must not contain repeated digits.

*There must not be any empty row.

*There must not be any row which is totally filled.

*There must not be any empty column.

*There must not be any column which is totally filled.

Here are examples of INVALID cases (according to the above conditions, respectively):


*

*An empty $3$-by-$3$ grid:





*

*A totally filled $3$-by-$3$ grid:





*

*Repeated digits in a $3$-by-$3$ grid:





*

*Repeated digits in a row:





*

*Repeated digits in a column:





*

*An empty row:





*

*A totally filled row:





*

*An empty column:





*

*A totally filled column:




Note that $C1,C2,C3,\dots,C9,R1,R2,R3,\dots,R9$ are positioned/attached to the $9$-by-$9$ grid to indicate that a rotated or transposed $9$-by-$9$ grid is considered as a different way (i.e. a rotated or transposed $9$-by-$9$ grid must be counted). Here are three examples of different VALID cases:
First example:

Second example:

Third example:


I saw similar question in this forum (sudoku problems), but those questions are not exact to this one.
Your help would be really appreciated. THANKS!
 A: We can improve the upper bound of the required number $N$ of ways from my comment as follows. 
Consider the first column of the filled grid. Assume it has $k$ zeroes. Then $1\le k\le 8$ and there are ${9\choose k}$ ways to place them in the column and $\tfrac {9!}{k!}$ ways to fill the remaining $9-k$ cells with positive numbers. 
Now consider the possibilities to fill a row. 
Assume that a row starts from a zero. It there are $m$ remaining zeroes in the row then $0\le m\le 7$ and there are ${8\choose m}$ ways to place them in the row and $\tfrac {9!}{(m+1)!}$ ways to fill the remaining $8-m$ cells with positive numbers. This give us in total $N_0=\sum_{m=0}^7 {8\choose m}\tfrac {9!}{(m+1)!}=4596552$ possibilities to fill a row.
Assume that a row starts from a positive number. It there are $m$ zeroes in the row then $1\le m\le 8$ and there are ${8\choose m}$ ways to place them in the row and $\tfrac {8!}{m!}$ ways to fill the remaining $8-m$ cells with positive numbers. This give us in total $N_+=\sum_{m=1}^8 {8\choose m}\tfrac {8!}{m!}=1401409$ possibilities to fill a row.
In total, we have a bound 
$N\le \sum_{k=1}^8 {9\choose k} \tfrac {9!}{k!} N_0^k N_+^{9-k}\approx 1.438\cdot 10^{64}$.
Again, this bound can be improved if we take into account that the rows are not independent but cannot contain a common positive number in the same column. But it seems this way leads to more complicted calculations and not a very big improvement.
