Different ways to place the workers on Santorini board game

Question

This combinatorics problem is based on the board game Santorini.

Consider a $5 \times 5$ chess board (but all squares are equal, say they're all white).

Player A places $2$ (equal) white workers in $2$ of the $25$ squares.

Player B then places $2$ (equal) black workers in $2$ of the remaining $23$ squares.

How many different ways can the players place their workers (including rotations and reflections)

If we exclude rotations and reflections, then the number is $\binom{25}{2}\times\binom{23}{2} = 75 900$.

But if we consider them this seems a lot more complicated... Is there an easy way to think about this?

What I mean by considering rotations and reflections is that the following configurations should be treated as the same:

"original configuration"

rotate $90$ degrees clockwise

rotate $180$ degrees

vertical axis reflection

diagonal axis reflection

Edit: If anyone who knows Burnside's Lemma could confirm my answer, I would appreciate it.

Answer 1

As said in one of the comments, using Burnside's lemma could be the way to go.

I've never used it before so this might not be completely correct. Please let me know if you find any mistake as I'm not completely sure of this answer.

Let $X$ be the set of $\binom{25}{2}\binom{23}{2}=75900 \ $ possible placements that can be done in one particular orientation.

Let $G$ be the rotation group on $X$.

$G$ has $8$ elements:

• the identity
• two $90^{\circ}$ edge rotations
• one $180^{\circ}$ edge rotation
• four reflections (we can see this as flipping the board $180^{\circ}$): two diagonal, one vertical and one horizontal

The solution of the problem can be found by counting the sizes of the fixed sets for the $8$ elements of $G$:

• the identity leaves all $75900$ elements of $X$ unchanged
• the two $90^{\circ}$ edge rotations leave $0$ elements of $X$ unchanged
• the $180^{\circ}$ edge rotation leaves $\binom{12}{2}\times 2 = 132$ elements of $X$ unchanged
• the four reflections leave $\binom{5}{4}\binom{4}{2}+\binom{5}{2}\times 10\times 2+\binom{10}{2}\times 2 = 320$ elements of $X$ unchanged

Thus, the total number of different configurations is $$\frac{1}{8}(1\times 75900+2\times 0+ 1 \times 132 + 4 \times 320) = 9664$$