everybody!Ahh.. I have some issues with this problem, I kind of struggle with it. Can somebody give me some help? I found that for a particular $n$ (I tried for 2,3,4,5) that there is just one sum. Though, I couldn't prove that in general.
We have p students. Each one have a board with $n$x$n$ squares ($n$ lines and $n$ columns, like a chess board). Each board is filled with the natural numbers from $1$ to $n^2$, in increasing order from the top to the bottom (first line contains the numbers $1,2,3,...,n$, the second contains the next $n$ numbers and so on). Now they have to put the sign minus in front of $mn$ numbers, such that on every line and every column there would be $m$ signs of minus. Each one need to calculate the sum of the numbers on their own board(the modified one). How many different sums can be obtained of those p students, knowing that all of them filled their boards correctly? Also, the numbers $n$ and $m$ are fixed and $m<n$.
HINT
Show that any kind of distribution of minus signs can be transformed into any other by repeating the following step:
If you have a minus sign in square $(i,j)$ (row $i$ and column $j$), and a minus sign in $(k,l)$ with $i \not = k$ and $j \not = l$, and you do not have a minus sign in either $(i,l)$ or $(k,j)$, then you can move the minus sign from $(i,j)$ to $(i,l)$, and the minus sign from $(k,l)$ to $(k,j)$
Note that any such $1$ step will keep the sum the same, so no matter how many of these kinds of steps we take, the sum remains the same as you transform one distribution into another. Also, to show that all distributions can be transformed into each other, it might be helpful to show that such distributions can be transformed to and from a kind of 'canonical' distribution, i.e one that has a 'nice' pattern of minus signs.