We know that n queens may always be placed on a square board of n²
squares (if n be greater than 3) without any queen attacking another
queen. But no general formula for enumerating the number of different
ways in which it may be done has yet been discovered; probably it is
undiscoverable. The known results are as follows:--
Where n = 4 there is 1 fundamental solution and 2 in all.
Where n = 5 there are 2 fundamental solutions and 10 in all.
Where n = 6 there is 1 fundamental solution and 4 in all.
Where n = 7 there are 6 fundamental solutions and 40 in all.
Where n = 8 there are 12 fundamental solutions and 92 in all.
Where n = 9 there are 46 fundamental solutions.
Where n = 10 there are 92 fundamental solutions.
Where n = 11 there are 341 fundamental solutions.
Obviously n rooks may be placed without attack on an n² board in n!
ways, but how many of these are fundamentally different I have only
worked out in the four cases where n equals 2, 3, 4, and 5. The answers
here are respectively 1, 2, 7, and 23. (See No. 296, "The Four Lions.")
We can place 2n-2 bishops on an n² board in 2^{n} ways. (See No. 299,
"Bishops in Convocation.") For boards containing 2, 3, 4, 5, 6, 7, 8
squares, on a side there are respectively 1, 2, 3, 6, 10, 20, 36
fundamentally different arrangements. Where n is odd there are
2^{½(n-1)} such arrangements, each giving 4 by reversals and
reflections, and 2^{n-3} - 2^{½(n-3)} giving 8. Where n is even there
are 2^{½(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3}
- 2^{½(n-4)}, each giving 8.
We can place ½(n²+1) knights on an n² board without attack, when n
is odd, in 1 fundamental way; and ½n² knights on an n² board, when
n is even, in 1 fundamental way. In the first case we place all the
knights on the same colour as the central square; in the second case we
place them all on black, or all on white, squares.