Considering Sudoku 4

There's something that doesn't make sense here, which I will explain with some examples. You fill in the seven squares in the upper-left, then the first column, then the first row. This gives you a count of $(9)_7 \cdot 6! \cdot 6!$ .

Suppose the diagram is as follows:

$\begin{array}{||c|c|c||c|c|c||c|c|c||} \hline \hline 1 & 4 & 7 & 2 & 5 & 8 & 3 & 6 & 9 \\ \hline 2 & 5 & & & & & & & \\ \hline 3 & 6 & & & & & & & \\ \hline \hline 4 & & & & & & & & \\ \hline 5 & & & & & & & & \\ \hline 6 & & & & & & & & \\ \hline \hline 7 & & & & & & & & \\ \hline 8 & & & & & & & & \\ \hline 9 & & & & & & & & \\ \hline \hline \end{array}$

There are 36 ways to complete the second column, as you state.

Now, consider another diagram, with 6 and 7 switched in the upper-left square:

$\begin{array}{||c|c|c||c|c|c||c|c|c||} \hline \hline 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 & 9 \\ \hline 2 & 5 & & & & & & & \\ \hline 3 & 7 & & & & & & & \\ \hline \hline 4 & & & & & & & & \\ \hline 5 & & & & & & & & \\ \hline 6 & & & & & & & & \\ \hline \hline 7 & & & & & & & & \\ \hline 8 & & & & & & & & \\ \hline 9 & & & & & & & & \\ \hline \hline \end{array}$

In the second column, the 6 must go in the bottom $3 \times 3$ square, and the 8 and 9 must go in the middle $3 \times 3$ square. This gives $3 \cdot 3 \cdot 2 = 18$ ways of placing these numbers. Then the remaining numbers 1, 2, 3 can be placed arbitrarily, for another 6 ways. Thus, for this diagram, there is a total of $18 \cdot 6 = 108$ ways of completing the second column.

We can't just multiply the number $(9)_7 \cdot 6! \cdot 6!$ by $36 + 108$ , or some other number, because among the diagrams counted in this count of $(9)_7 \cdot 6! \cdot 6!$ , some have 36 ways of completing the second column and others have 108 ways. We have to count the number of diagrams of each kind, and then multiply accordingly.

Here is how I counted. There are 9! ways of filling the first column. Then I'm going to divide into cases similar to yours.

Case 1 : b1, b2, b3 are the same as a4, a5, a6, in some order.

There are 3! ways to order b1, b2, b3. Then there are 36 ways to complete the second column.

Case 2 : Of the values b1, b2, b3, one comes from a4, a5, a6, and two come from a7, a8, a9.

There are $3 \cdot 3 = 9$ ways to choose b1, b2, b3, then 3! ways to order them. Then there are 108 ways to complete the second colun.

Case 3 : Of the values b1, b2, b3, two come from a4, a5, a6, and one comes from a7, a8, a9.

This is the same as Case 2.

Case 4 : b1, b2, b3 are the same as a7, a8, a9, in some order.

This is the same as Case 1.

Thus, the number of ways of filling the second column is $6 \cdot 36 + 9 \cdot 6 \cdot 108 + 9 \cdot 6 \cdot 108 + 6 \cdot 36 = 12096$ .

There are then 3 ways to fill in the seventh cell of the upper-left square, then 6! ways to fill the rest of the first row. This gives us a final count of $9! \cdot 12096 \cdot 3 \cdot 6! = 9 \, 481 \, 096 \, 396 \, 800.$