Problem on arrangement of numbers in 4×4 array Problem:-
How many different 4×4 arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
I tried to Fix 1,-1 on each column so that sum of each column is zero but then sum of some rows aren't zero
I'm confused how to arrange them in order to sum zero,I shall be thankful if somebody explain me the solution since i'm weak in combinatorics
 A: It's the same thing as creating $(4\times4)$-matrices $A=[a_{ij}]$ with two $+$ and two $-$ in each row and each column. 
Assume $a_{11}=+$. Here we have chosen $1$ of $2$ possible cases. Makes $2$ ways.
Then we need one more $+$ in the first row and one more $+$ in the first column. Assume $a_{12}=a_{21}=+$. Here we have chosen $1$ of $3\times3=9$ possible cases. Makes $9$ ways.


*

*When $a_{22}=+$ as well the rest of $A$  is determined. Makes $1$ way.

*When $a_{22}=-$ then we need one more $+$ in the second row and one more $+$ in the second column. Assume $a_{23}=a_{32}=+$. Here we have chosen $1$ of $2\times2=4$ possible cases. The rest of $A$ is then determined. Makes $4$ ways.


It follows that there are
$$2\cdot9\cdot(1+4)=90$$
admissible matrices. 
A: For the first row, we have to assign two $1$s en two $-1$s, for a total of $4 \choose 2$ combinations. Then, we can choose one of three options:


*

*Repeat all values, so that the remaining two rows are defined;

*Change all values, so that the sum of the top two values in each column equals $0$. In this case, the third row can be chosen again in $4 \choose 2$ ways, after which the remaining row is defined;

*Change two out of four values with a different sign (if they had the same sign, the sum of the second row would equal $4$ or $-4$). This can happen in ${2 \choose 1}{2 \choose 1}$ ways, after which the corresponding two columns are defined. Two possibilities for the last four elements remain, so we have to multiply by $2 \choose 1$ again.
Overall, the number of possible arrays thus equals:
$${4 \choose 2} \left(1 + {4 \choose 2} + {2 \choose 1}{2 \choose 1}{2 \choose 1} \right) = 6 \cdot (1 + 6 + 8) = 90$$
