In how many ways can the numbers 1 through9 be placed once each in the squares below, if the numbers 1 and 2 cannot be placed in adjacent squares (either horizontally, vertically, or diagonally), and rotations and reflections of the grid are considered the same?
The squares in the grid are all the same size and it is a 3x3 grid (3 rows and 3 columns)
Only possible way I think of is to fill in the squares with numbers and see if they work sort of a trial and error
+128399 ____ _____ _____
1 2 3
_____ _____ ______
4 5 6
_____ _____ ______
7 8 9
If the "1" is placed in any of the 4 corner positions
For each of these, the "2" can occupy any 5 positions and the other 7 digits can occupy the other 7 positions in 7! ways
This gives us 4 * 5 * 7! = 100800 ways
If the "1" is placed in positions "4" or "6" the "2" can only occupy only 3 positions and the other digits can be arranged in 7! ways = 2 * 3 * 7! = 30240 ways
If the "1" is placed in positions "2" or "8" the "2" can only occupy 3 positions and the other 7 digits can occupy the other 7 positions in 7! ways = 2 * 3 * 7! = 30240 ways
Finally.....the "1" cannot be placed in the middle position because the "2" will have to be next to it in some way
So....the total ways = 100800 + 2(30240) = 161280 ways
+2666 I think it has to do with the reflection and rotation part...
1 2 3
For example, with Chris's method, he counts for each of these below, which can all be reflected and or rotated to get the original grid (4 5 6)
7 8 9
1 2 3 7 4 1 9 8 7 3 6 9
4 5 6 8 5 2 6 5 4 2 5 8
7 8 9 9 6 3 3 2 1 1 4 7
3 2 1 9 6 3 7 8 9 9 6 3
6 5 4 8 5 2 4 5 6 8 5 2
9 8 7 7 4 1 1 2 3 7 4 1
7 8 9 1 4 7 3 2 1 1 4 7
4 5 6 2 5 8 6 5 4 2 5 8
1 2 3 3 6 9 9 8 7 3 6 9
9 8 7 3 6 9 1 2 3 7 4 1
6 5 4 2 5 8 4 5 6 8 5 2
3 2 1 1 4 7 7 8 9 9 6 3
There are 4 ways to rotate the square, and 4 ways to reflect it, so i think it is \(161280 \div 16 = \color{brown}\boxed{10080}\) ways
