## THE SIX PAWNS.

(

The Guarded Chessboard)

In how many different ways may I place six pawns on the chessboard so

that there shall be an even number of unoccupied squares in every row

and every column? We are not here considering the diagonals at all, and

every different six squares occupied makes a different solution, so we

have not to exclude reversals or reflections.

## Answer:

The general formula for six pawns on all squares greater than 2 squared 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 x 7 x 6)/(1 x 2 x 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 x 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 x 6 = 18,816,

which is the total number of ways, as we have already stated.