<s>[INST] Put into each cell one number 1 or one number -1 that the sums of number in each coloums each rows is 0. There is a 4 x 4 grid. In each cell of the grid, you are allowed to put either a 1 or a -1. The sum of the numbers of each column and each row must equal 0. How many such configurations of the grid are there?
 [/INST] You can to start completing the first line of the square
\begin{array}{|c|c|c|c|}
\hline
 .& . & .&. \\ \hline
 .& .&. & .\\ \hline
. & . & .& .\\ \hline
 .& . & .& .\\ \hline 
\end{array}
And you can do this of $ 6 = \binom{4}{2} $ ways because you need to choose the 2 places between the 4 to put the numbers 1. For exemplo, 
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 & -1 \\ \hline
. & .&.& .\\ \hline
. & .& .&.\\ \hline
 .&.&. &.\\ \hline
\end{array}. 
Then you can complete the fisrt column choosing one of the three places to put the number 1 . For exemplo, 
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & & &\\ \hline
. & . & .&.\\ \hline
 .&  .& .&.\\ \hline
\end{array}. 
Then, after you do this, your table must be
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & & &\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}. 
You can complete the column that contains the number 1. There are two signicant ways to do this. You can to put the 1 in the column that already countains the 1 or no.  These cases are respectively of this forms
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}.
or for examplo
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}.
Notice  that there are another way to do the second case, namely
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1& -1&1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}.
The case \begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}
is determined to be 
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & -1&1 &-1\\ \hline
-1 &1  &-1 &1\\ \hline
-1 &1  &-1 &1\\ \hline
\end{array}.
And the second case 
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 &- 1&-1 &1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}
and
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 &  & &\\ \hline
-1 &  & &\\ \hline
\end{array}.
also are to be determined by respectvely
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 &- 1&-1 &1\\ \hline
-1 & 1 & &\\ \hline
-1 & 1 & &\\ \hline
\end{array}
and
\begin{array}{|c|c|c|c|}
\hline
1& -1 & 1 &-1 \\ \hline
1 & 1& -1&-1\\ \hline
-1 &  & &1\\ \hline
-1 &  & &1\\ \hline
\end{array}.
Then you need to choose the polace where you put the number 1 on the third and this determines the table. This can be done of 2 ways. Then the answer is
$$ 6 \cdot 3(1 + 2 \cdot 2) = 90 \quad \mbox{ways.}  $$
</s>