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 2n 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 2½(n-1) such arrangements, each giving 4 by reversals and reflections, and 2n-3 - 2½(n-3) giving 8. Where n is even there are 2½(n-2), each giving 4 by reversals and reflections, and 2n-3 - 2½(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.
THE TWO PIECES PROBLEM.
On a board of n2 squares, two queens, two rooks, two bishops, or two knights can always be placed, irrespective of attack or not, in ½(n4 - n2) ways. The following formulæ will show in how many of these ways the two pieces may be placed with attack and without:—
| With Attack. | Without Attack. | |
| 2 Queens | 5n3 - 6n2 + n | 3n4 - 10n3 + 9n2 - 2n |
| 3 | 6 | |
| 2 Rooks | n3 - n2 | n4 - 2n3 + n2 |
| 2 | ||
| 2 Bishops | 4n3 - 6n2 + 2n | 3n4 - 4n3 + 3n2 - 2n |
| 6 | 6 | |
| 2 Knights | 4n2 - 12n + 8 | n4 - 9n2 + 24n |
| 2 |