I mean that for each choice of the first two rows that is not the same, there will be 4 ways to fill out the third row. For example, if the first row is 1, 1, -1, -1 and the second row is 1, -1, 1, -1, see how many ways you can fill out the 3rd row.
There are six possible strings that can be used as rows in a \(4\times 4\) grid.
\(\begin{array}{*{20}{c}}
{1}& + & + & - & - \\
{2}& - & - & + & + \\
{3}& + & - & - & + \\
{4}& - & + & + & - \\
{5}& + & - & + & - \\
{6}& - & + & - & + \end{array}\)
Note that the pairs \(1,2\;\;3,4~\&\;~5,6\) are compliments( negatives) of each other.
Also that both \(3,4\) are palindromes. which complicates over counting.
Consider the \(4\times 4\) grid made of rows \(1,2,3,4\) that is a
successful grid because each row & each column contain two pluses and two minuses. Moreover the rows as a whole can be rearranged in \(4!=24\) ways.
At this point we might be temped to say well just take \(\mathcal{C}_4^6=15\) grids.
But wait! The \(4\)-tuple \((1,3,5,6) \)is not a successful grid because the first column has three pluses.
Interestingly enough the \(4\)-tuples \((1,2,3,4),\;(1,2,5,6)\;\&\;(3,4,5,6 \) are all a successful grid.
Again wait. The \(4\)-tuple \((1,1,2,2) \)is also a successful grid.
BUT how many
different ways can it be rearranged?