What is the maximum amount of checkers you can place on an $m \times n$ checkerboard such that no four checkers make a rectangle parallel to the rows and columns? Does there exist a closed form for the relationship between the number of checkers and the size of the checkerboard?
There are really too many cases and one specific solution for 4 checkers is below.
The number of possible rectangles is given by the number of ways to choose two columns times the number of ways to choose two rows. i.e.: $$\frac{nm(n-1)(m-1)}{4}$$ The number of ways to put $4$ checkers on the board is: $${nm}\choose{4}$$ Thus, the number of ways to put $4$ checkers on a board without rectangles is just: $${{nm}\choose{4}} - \frac{nm(n-1)(m-1)}{4}$$ This does not answer your question regarding $k$ checkers on the board
So, what is the generalization of this question? And does there exist a closed form? And how to apply some counting skills in this problem and maybe apply some knowledge in the combinatorics design like my duplicated questions?
An upper bound on $k$.
Let $X_i$ be the number of checkers placed in row $i$, so we have $\sum\limits_{i=1}^n X_i=k$.
We clearly need $\sum\limits_{i=1}^n \binom{X_i}{2}\leq \binom{m}{2}$.
If $k=qn+r$ then the sum on the left is at least $(n-r)\binom{q}{2}+r\binom{q+1}{2}$.
This allows you to find a bound on $k$, I don't know when this bound is sharp though, it is sharp in the case in which a $(m,k/n,3)$-design exists. And possibly in some other cases, in which the largest value of $k$ that satisfies $(n-r)\binom{q}{2}+r\binom{q+1}{2}$ gives us a value strictly larger than $\binom{m}{2}$.