The number of different ways is 63,504. The general formula for such
arrangements, when the number of letters in the sentence is 2_n_ + 1, and
it is a palindrome without diagonal readings, is [4(2^_n_ - 1)]^2.
I think it will be well to give here a formula for the general solution
of each of the four most common forms of the diamond-letter puzzle. By
the word "line" I mean the complete diagonal. Thus in A, B, C, and D, the
lines respectively contain 5, 5, 7, and 9 letters. A has a non-palindrome
line (the word being BOY), and the general solution for such cases, where
the line contains 2_n_ + 1 letters, is 4(2^_n_ - 1). Where the line is a
single palindrome, with its middle letter in the centre, as in B, the
general formula is [4(2^_n_ - 1)]^{2}. This is the form of the
Rat-catcher's Puzzle, and therefore the expression that I have given
above. In cases C and D we have double palindromes, but these two
represent very different types. In C, where the line contains 4^n-1
letters, the general expression is 4^(2^{2_n_}-2). But D is by far the
most difficult case of all.
I had better here state that in the diamonds under consideration (i.) no
diagonal readings are allowed--these have to be dealt with specially in
cases where they are possible and admitted; (ii.) readings may start
anywhere; (iii.) readings may go backwards and forwards, using letters
more than once in a single reading, but not the same letter twice in
immediate succession. This last condition will be understood if the
reader glances at C, where it is impossible to go forwards and backwards
in a reading without repeating the first O touched--a proceeding which I
have said is not allowed. In the case D it is very different, and this is
what accounts for its greater difficulty. The formula for D is this:
[Illustration:
(_n_+5)x2^{2_n_+2} + (2^{_n_+2}x(1x3x5x7
. . . . . (2n-1)) / _n_) - 2^{_n_+4} - 8 ]
where the number of letters in the line is 4_n_+1. In the example given
there are therefore 400 readings for _n_ = 2.