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"Different Dispositions in the ChessBoard" by Timothy CHOW, Mario VELUCCHI
(16 Nov. 1995)


PROBLEM:
How many different dispositions I have in an order-N chessboard, if I place K Queens?

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The answer is in the "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
... ...             ...       ...          . ...
... ...             ...       ...          . ...


Mario VELUCCHI
Via Emilia, 106
I-56121 Pisa - ITALY

e-mail: velucchi@cli.di.unipi.it