I'm trying to get my head around a problem, and it's not working.

The problem: consider an NxN matrix that represents a binary number. For instance, a 4x4 matrix is a 16 bit number, a 6x6 matrix is a 36 bit number. N will always be even.

Now, for any given N, how many matrices / numbers are there where every row and column has exactly half of its values 0 and half 1?

Here is what I have learned so far:

  • From brute force, I know that the answer for a 2x2 matrix is 2. For a 4x4 matrix, the answer is 90.
  • I can calculate the number of rows OR columns that half half of their bits on with the formula: $$ f(n) = {n! \over (n/2)!^2} * 2 ^ {n^2} $$
  • But I can't simply take percentage of rows times percentage of columns, presumably because they are not independent.

This started out of curiosity and has become an obsession. Please help.

  • $\begingroup$ nice question! +1 $\endgroup$ – sigmatau Jul 23 '13 at 2:32

This sequence appears in OEIS, wherein there are references such as this article in the Electronic Journal of Combinatorics, which gives asymptotic results.

The sequence, incidentally, begins $1, 2, 90, 297200, 116963796250, 6736218287430460752,\ldots$

  • $\begingroup$ Thanks for that -- it is indeed the correct sequence. And the article is interesting. So far, it looks like there is no general solution, which is pretty frustrating. But thanks for the additional info! $\endgroup$ – user977705 Jul 23 '13 at 14:40

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