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Placing checkers on an m x n board This problem seems to relate to the combinatoric design which is unfamiliar for me. Still, it arises my interests.

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? Is there existed a close 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 is there existed a close form? and how to apply some counting skills in this problem and maybe applied some knowledge in the combinatorics design like my duplicated questions.

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    I'm going to recommend that this Question be merged with the "possible duplicate" you identified (apparently by a different OP), thus preserving the new Answer. Even the body of this Question is largely a quotation of the Answer on the old Question [posted by nbubis](http://math.stackexchange.com/a/2093180/3111), which deserves attribution. – 2017-01-24
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    you promised that you would continue to answer. – 2017-01-25
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    I'm happy to do more with my answer and to comment on answers by others. But I'm confused by the two açcounts -- are you the same as the one who asked this before? – 2017-01-25
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    absolutely not, I was another person – 2017-01-25
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    If you'd like do more answer, I'll be very happy – 2017-01-25
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    @hardmath, would you like to do some further work on this question, I'm particularly interested in this question and after few days work, I still have no idea. – 2017-01-25
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    Ok, one point I was hinting at before was that it's a lot to expect any "simple" closed form because if we knew the answer to maximum checkers on a board, we'd also know the answer to a classic open problem in combinatorics, the projective plane [Prime Power Conjecture](https://symomega.wordpress.com/2010/02/26/open-problems-in-finite-geometry-the-prime-power-conjecture/). – 2017-01-25
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    thanks if you can answer my question even just some continued hints – 2017-01-25

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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 $\binom{m}{2}$.

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    what is this mean (m,w/n,3) design, and can you give me some detailed explanation, thank you. – 2017-01-24
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    Because I am not very familiar with combinatorics design. – 2017-01-24
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    When your $w$ (the $m$ of Question statement) is at least $\binom{n}{2}$, then the maximum number of checkers is $\binom{n}{2} + m$ by having $\binom{n}{2}$ rows with two checkers and the rest of the rows one checker apiece. – 2017-01-26
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    I have no idea why I wrote a $w$ variable, I never use it. My $m$ is the same $m$ as in the problem. – 2017-01-26
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    @hardmath thank you${}{}{}{}$ – 2017-01-26
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    @hardmath would you like give me a formal proof that there is no expression existed. – 2017-03-07