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Find the number of $4 \times 4$ matrices such that its elements are $1$ and $-1$ and the sum of elements in each of its rows and columns is $0$.
I am only able to make trivial progress that there are $6$ ways to fill a row and $3$ ways to fill a column.I couldn't find a way to proceed further.
Can this be generalised to a $n \times n$ matrix under the same conditions?
For the $4\times 4$ case, the number of ways is $90$ and the method is as follows:
The two $+$ in the first row can be placed in six ways. Take the first row to be $++--$; we will have to multiply the counting by six in the end.
Then the second row can be formed in three essentially different ways.
(a) $++--$: Now the third and fourth rows are completely determined; there is one such matrix.
(b) $--++$: Now the third row is completely arbitrary (the two $+$ can be in any positions) giving $6$ possibiliites; but then the fourth row will be determined.
(c) $+-+-$: There are four "equivalent" ways of doing this for the second row, The other three are $+--+$, $-++-$, and $-+-+$; consider just the first one $+-+1$. Then The items in the first and fourth coumn in the third and fourth rows are completely determined, and the little $2\times 2$ square remaining has two allowable arrangements (the $(3,2)$ element can be $+$ or $-$, after which the others are determined). So case (c) gives $4\times 2=8$ allowed arrangements.
The total answer is $$ 6 (1+6 + 4\cdot 2) = 90 $$ ADDED
I am working on the general case of $2m \times 2n$.
For $2m \times 2$ the answer is of course $\binom{2n}{n}$.
For a $6\times 4$ matrix the answer appears to be $$ 20(1\cdot 1 + 9 \cdot 2 + 9 \cdot 6 + 1 \cdot 20) = 1860 = 31\cdot 60$$
The $4\times 2n$ cases
For $m=2$, the number of allowable $4\times 2n$ matrices is obtained by asking how many ($k$) minus signs in the second row lie under $+$ signs in the first row. The number of arrangements of the non-fixed elements of the third and 4-th row is of the form $\binom{2k}{k}$. You can get the expression in closed form: $$ N(4\times 2n) = \binom{2n}{n}\sum_{k=0}^n \binom{n}{k}^2\binom{2k}{k} = \binom{2n}{n} F\left( \left. \begin{array}{c} \frac12, -n, -n\\1,1 \end{array}\right| 4\right) $$ where the $F$ is a hypergeomentric function of three upper and two lower indices (sometimes written _3F2) as specified above.
$$ N(4,2) = 3\binom{2}{1}\\ N(4,4) = 15\binom{4}{2}\\ N(4,6) = 93\binom{6}{3}\\ N(4,8) = 939\binom{8}{4}\\ N(4,10) = 4653\binom{10}{5}\\ N(4,12) = 35169\binom{12}{6}\\ \vdots $$
The $6\times 2n$ cases
Counting the allowable $6\times 2n$ matrices is more difficult, since a column of the $4 \times 2n$ matrices underneath the first two rows are not fixed when the first two rows contain the same symbol in that column. I don't have, at this point, a good way of attacking this.
