How many different ways (not counting rotations and reflections of the board as different) can K queens be...
competition/games/chess/queen.placements New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
How many different ways (not counting rotations and reflections of the board as different) can K queens be placed on an NxN chessboard? [ Editor: problem statement reconstructed from the postings, which ask for the number of "different dispositions"; the formula below counts placements up to the eight symmetries of the board. ]
Show solution
The answer is in "Polya's Theorem".
The wanted number is 1/8th of the coefficient of
a^K * b^(N^2-K) in these polynomials:
if N is even
p(a,b,N) := (a+b)^(N^2) + 2*(a+b)^N*(a^2+b^2)^((N^2-N)/2)
+ 3*(a^2+b^2)^(N^2/2) + 2*(a^4+b^4)^(N^2/4)
if N is odd
p(a,b,N) := (a+b)^(N^2) + 2*(a+b)*(a^4+b^4)^((N^2-1)/4)
+ (a+b)*(a^2+b^2)^((N^2-1)/2) + 4*(a+b)^N*(a^2+b^2)^((N^2-N)/2)
Note:
the total number of dispositions is equal to: Binomial coefficient(N^2, K)
and the number of different dispositions is near 1/8th.
Samples (K = N):
for N = 1 different dispositions = 1
for N = 2 different dispositions = 2
for N = 3 different dispositions = 16
for N = 4 different dispositions = 252
for N = 5 different dispositions = 6814
for N = 6 different dispositions = 244344
for N = 7 different dispositions = 10746377
for N = 8 different dispositions = 553319048
for N = 9 different dispositions = 32611596056
for N = 10 different dispositions = 2163792255680
...
for N = 14 different dispositions = 110066314584030859544
for N = 15 different dispositions = 11375695977099383509351
...
-- Timothy Chow and Mario Velucchi <velucchi@CLI.DI.Unipi.IT>
[ Editor: an earlier version of this message gave 5532069048 for N = 8;
the corrected value 553319048 is used above. It checks: with K = N = 8
the four terms give C(64,8) = 4426165368, 2*139448, 3*C(32,4) = 107880
and 2*C(16,2) = 240, whose sum 4426552384 divided by 8 is 553319048. ]