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?

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
$$ 6 \cdot 3(1 + 2 \cdot 2) = 90 \quad \mbox{ways.}  $$
