**THE SIX PAWNS.— solution**

The general formula for six pawns on all squares greater than 2^{2}
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.

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.

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