11.5. In how many ways can a table of size $n \times n$ cells be filled with zeros and ones so that each row and each column contains an even number of ones? Each cell of the table must contain either a zero or a one.
Answer: $2^{(n-1)^{2}}$.

Solution. We will prove that any arrangement of zeros and ones in the cells of the lower left corner square of size $(n-1) \times(n-1)$ can be uniquely extended to fill the entire table with zeros and ones, satisfying the conditions of the problem. Consider any of the lower $n-1$ rows of the table. If there is an even number of zeros in the left $n-1$ cells, then according to the condition, a 0 must be placed in the remaining right cell, and if there is an odd number of zeros, then a 1 must be placed. This is always possible to do, and in a unique way. Similarly, the upper cells of the left $n-1$ columns of the table are uniquely filled. Note that the parity of the number of ones written in the right column coincides with the total number of ones in the cells of the lower left corner square of size $(n-1) \times(n-1)$ and coincides with the parity of the number of ones written in the top row of the table. Therefore, to satisfy the condition of the problem, a 0 must be placed in the top right cell of the table if all these three numbers are even, and a 1 in the opposite case. This is also always possible and is done uniquely. From the above, it follows that the desired number of ways is equal to the number of all ways to fill a table of size $(n-1) \times(n-1)$ cells with zeros and ones, which is, obviously, $2^{(n-1)^{2}}$.

Criteria. The idea that any arrangement of zeros and ones in the cells of the lower left corner square of size $(n-1) \times(n-1)$ can be uniquely extended to fill the entire table with zeros and ones, satisfying the conditions of the problem: 1 point. The same idea with a proof: 5 points. Correct calculation of the desired number of ways to arrange after this: 2 points.