How many ways to fill the $4 \times 4$ board by nonnegative integers, such that sum of the numbers of each row and each column is $3$?
I wrote a brute-force and got $2008$ which seems to be the right answer. I was wondering what is the combinatorial way of solving this one? How about the general problem of $N\times N$ or $N\times K$ with sum a $R$?