The number of ways in which the four pigs may be placed in the thirty-six
sties in accordance with the conditions is seventeen, including the
example that I gave, not counting the reversals and reflections of these
arrangements as different. Jaenisch, in his _Analyse Mathematique au jeu
des Echecs_ (1862), quotes the statement that there are just twenty-one
solutions to the little problem on which this puzzle is based. As I had
myself only recorded seventeen, I examined the matter again, and found
that he was in error, and, doubtless, had mistaken reversals for
different arrangements.
Here are the seventeen answers. The figures indicate the rows, and their
positions show the columns. Thus, 104603 means that we place a pig in the
first row of the _first_ column, in no row of the _second_ column, in the
fourth row of the _third_ column, in the sixth row of the _fourth_
column, in no row of the _fifth_ column, and in the third row of the
_sixth_ column. The arrangement E is that which I gave in diagram form:--
A. 104603
B. 136002
C. 140502
D. 140520
E. 160025
F. 160304
G. 201405
H. 201605
I. 205104
J. 206104
K. 241005
L. 250014
M. 250630
N. 260015
O. 261005
P. 261040
Q. 306104
It will be found that forms N and Q are semi-symmetrical with regard to
the centre, and therefore give only two arrangements each by reversal and
reflection; that form H is quarter-symmetrical, and gives only four
arrangements; while all the fourteen others yield by reversal and
reflection eight arrangements each. Therefore the pigs may be placed in
(2 x 2) + (4 x 1) + (8 x 14) = 120 different ways by reversing and
reflecting all the seventeen forms.
Three pigs alone may be placed so that every sty is in line with a pig,
provided that the pigs are not forbidden to be in line with one another;
but there is only one way of doing it (if we do not count reversals as
different), as follows: 105030.