Solution
Two Tokens on a Chessboard
Show the problem again
Two identical tokens are placed on two different squares of an 8 × 8 board. Two placements are considered the same if one can be carried to the other by a symmetry of the square board (any of its 4 rotations or 4 reflections). How many different placements are there? Give an integer.
Worked solution
The answer is 278. Apply Burnside to the 8 symmetries acting on the $\binom{64}{2} = 2016$ placements. The 90° and 270° rotations split the squares into 4-cycles (no square is fixed on an even board), so no 2-set is fixed. The 180° rotation fixes a 2-set exactly when it is a pair $\{p, p'\}$ of centrally opposite squares: 32. The reflections in the two midlines fix no squares and pair them into 32 swapped pairs, so each fixes 32 placements. Each diagonal reflection fixes the 8 diagonal squares, so it fixes $\binom{8}{2} = 28$ placements with both tokens on the diagonal, plus $56/2 = 28$ swapped pairs: 56. Total $2016 + 0 + 0 + 32 + 32 + 32 + 56 + 56 = 2224$, and $2224/8 = 278$.
Source: Standard application of Burnside's lemma; no traceable original source. Statement written for AxiomIQ.