Number of Sudoku solutions with just the row and block rule

Each block is constrained by the two blocks sharing the same rows, so we consider a group of three blocks in isolation and cube this.

Consider filling the blocks sequentially. The first can be filled in 9! ways:

Now consider placing 4 5 and 6 in the second block. For the minute we'll just think about which rows they can go in.

There are two choices:

Choice 1

4, 5 and 6 lie in the same row eg.

But we could equally well have placed 4, 5 and 6 in the bottom row so this can be done in two ways.

Using each of these placements, the rows in which numbers can be placed in the third block are fixed, eg.

Choice 2

The second choice is to split 4,5 and 6 up: for example

There are 2 x ³C₂ = 6 ways of splitting 4, 5 and 6

For any given split there are 3 x 3 = 9 permitted arrangements of the remaining numbers (7 could have been 8 or 9 abd 3 could have been 1 or 2).

So that's 54 arrangements in total

Using each of these placements, the rows in which numbers can be placed in the third block are fixed again, eg.

Finally then, consider putting the columns back in (but not the column constraints yet).

There are 3! one of the 6 sets of 3 numbers into 3 columns and therefore we have: C = [9!*(2+6*3*3)*3!⁶]³