Here is a cross reference to a thread with the same question. Haven't looked over the details. Maybe you forgot you'd asked the question already?
Hooray! The forum is back up!
In how many different ways can you arrange the numbers 1-9 in a 3X9 grid such that each number appears exactly once in each row: and each number appears exactly once in each of the left, middle, and right 3X3 grids? Here is the grid along with one possible arrangement:
2 7 1 3 5 9 6 4 8
4 3 8 6 7 2 1 5 9
5 6 9 1 4 8 2 3 7
Here is what I am thinking: There are 9! ways to arrange the first row; then for the second row, first three columns we have 6 choose 3 ways to choose the numbers times 3! ways to arrange them. Now for the second row column four we have two cases:
If the number appearing in column four row one was used in row 2 columns one through three, we have 6 choices for row 2 column 4, otherwise we have 5 choices. If the number appearing in row one column 5 appeared in the first three columns of row 2, then we have 5 choices for row 2 column 5, otherwise we have 4 choices. The same for column six, we have either 4 choices or 3. For the final three columns we have 3! ways to arrange the final three numbers. The final row should have 3*3! ways to arrange the numbers.
This seems a little overly complicated. Can anyone critique my logic and maybe come up with a better approach?