The Monk might have placed dogs in the kennels in
hundred and twenty-six different ways, so that there should be ten dogs
on every side.
The number of dogs might vary from twenty to forty, and
as long as the Monk kept his animals within these limits the thing was
The general solution to this puzzle is difficult.
I find that for n
dogs on every side of the square, the number of different ways is (n4
+ 10n3 + 38n2
+ 62n + 33) / 48, where n is
odd, and ((n4 +
10n3 + 38n2
+ 68n) / 48) + 1, where n is
even, if we count only
those arrangements that are fundamentally different.
But if we count
reversals and reflections as different, as the Monk himself did, then n
dogs (odd or even) may be placed in ((n4
+ 6n3 + 14n2
+ 15n) /
6) + 1 ways.
In order that there may be n
every side, the
number must not be less than 2n nor greater than 4n,
but it may be
any number within these limits.