I was working on a Riemannian geometry research problem and I faced with an algebraic subproblem that can be described as follows:
I have a collection of $r$ balls on a box and two positive integers, $M$ and $N$. For each choice of $k$ of these $n$ balls, I write down two numbers $m, n$ where $m \leq M$ and $n \leq N$ and then I turn the ball's taken to the box. I assume that the balls are all distinguished and if on another draw I take the same balls, the numbers written must be the same. The rule is that I can never write $M$ and $N$ in the same $k$-ball draw. That is, if I write $M$, then the other number must be smaller than $N$, and vice versa. If you need to, you can assume $M> N$, or vice versa.
The question is: for which real numbers $A, B$ with negative $A$ or $B$ (if one is negative, the other has to be positive) and for which choices of the numbers to be written at each draw we have $$Am + Bn> 0,$$ for every pair $(m, n)$ annotated and $$ AM + BN <0.$$
I'm looking for algebraic relations between A and B and even and also, I'm trying to understand how restrictive $M$ and $N$ are and how restrictive the choice of numbers I can write down.
I'm making some assumptions in answering that I think match the spirit of the problem...apologies if I've taken too many liberties. Without loss of generality, we can let $B=-1$, since $A$ can scale appropriately. I'm going to assume that $m,n$ are positive integers.
Since there are $r$ balls, this problem boils down to picking $A$ so that there are at least $r \choose k$ solutions $(m,n)$ to $Am-n > 0$, but $AM-N < 0$ or equivalently $A < \frac{N}{M}$. For a fixed value of $n$, $Am-n > 0$ as long as $m > \left\lceil \frac{n}{A} \right\rceil$, so there are $\displaystyle \sum_{n=1}^N M-\left\lceil \frac{n}{A} \right\rceil$ such solutions, which must be greater than or equal to $r \choose k$.
The sum with the ceilings can get messy if you need very precise estimates, but if you can handle an approximation, $A$ will meet this condition if $A$ satisfies $$\begin{eqnarray*}{r \choose k}&\le&\sum_{n=1}^N M-\frac{n}{A}-1\\&\le&N(M-1)-\frac{N(N+1)}{2A}\end{eqnarray*}$$
Thus a sufficient condition for $A$ to satisfiy the conditions of the problem is $$\frac{N(N+1)}{2(N(M-1)-{r \choose k})} \le A < \frac{N}{M}$$