The general formula for six pawns on all squares greater than 2² is
this: Six times the square of the number of combinations of n things
taken three at a time, where n represents the number of squares on the
side of the board. Of course, where n is even the unoccupied squares
in the rows and columns will be even, and where n is odd the number of
squares will be odd. Here n is 8, so the answer is 18,816 different
ways. This is "The Dyer's Puzzle" (_Canterbury Puzzles_, No. 27) in
another form. I repeat it here in order to explain a method of solving
that will be readily grasped by the novice. First of all, it is evident
that if we put a pawn on any line, we must put a second one in that line
in order that the remainder may be even in number. We cannot put four or
six in any row without making it impossible to get an even number in all
the columns interfered with. We have, therefore, to put two pawns in
each of three rows and in each of three columns. Now, there are just six
schemes or arrangements that fulfil these conditions, and these are
shown in Diagrams A to F, inclusive, on next page.
[Illustration]
I will just remark in passing that A and B are the only distinctive
arrangements, because, if you give A a quarter-turn, you get F; and if
you give B three quarter-turns in the direction that a clock hand
moves, you will get successively C, D, and E. No matter how you may
place your six pawns, if you have complied with the conditions of the
puzzle they will fall under one of these arrangements. Of course it
will be understood that mere expansions do not destroy the essential
character of the arrangements. Thus G is only an expansion of form A.
The solution therefore consists in finding the number of these
expansions. Supposing we confine our operations to the first three
rows, as in G, then with the pairs a and b placed in the first and
second columns the pair c may be disposed in any one of the remaining
six columns, and so give six solutions. Now slide pair b into the
third column, and there are five possible positions for c. Slide b
into the fourth column, and c may produce four new solutions. And so
on, until (still leaving a in the first column) you have b in the
seventh column, and there is only one place for c--in the eighth
column. Then you may put a in the second column, b in the third, and c
in the fourth, and start sliding c and b as before for another series
of solutions.
We find thus that, by using form A alone and confining our operations to
the three top rows, we get as many answers as there are combinations of
8 things taken 3 at a time. This is (8 × 7 × 6)/(1 × 2 × 3) = 56. And it
will at once strike the reader that if there are 56 different ways of
electing the columns, there must be for each of these ways just 56 ways
of selecting the rows, for we may simultaneously work that "sliding"
process downwards to the very bottom in exactly the same way as we have
worked from left to right. Therefore the total number of ways in which
form A may be applied is 56 × 6 = 3,136. But there are, as we have seen,
six arrangements, and we have only dealt with one of these, A. We must,
therefore, multiply this result by 6, which gives us 3,136 × 6 = 18,816,
which is the total number of ways, as we have already stated.