Consider a $h \times l$ binary matrix (a matrix with all entries $a_{ij}$, $1 \le i \le h$, $1 \le j \le l$, equal to $0$ or $1$). We know that each row has $n \lt l$ entries equal to $1$ and $l - n$ entries equal to $0$. All rows are different. There are a total of ${l \choose 2}$ possible couples of columns in it. Let $c$ be the number of couples of columns with index $j$ and $k$ such that $a_{ij} = 1 \lor a_{ik} = 1$, $1 \le i \le h$. It's like there is a path of ones from the top to the bottom of the matrix along the two columns.
I would like to find a function $f(h, l, n)$ to have a lower bound for $c$:
$$c \ge f(h, l, n)$$
My thoughts: we have a total of $n \cdot h$ entries equal to $1$, thus on average a column will have $\frac{nh}{l}$ ones, therefore there exist a column with at least $\lceil\frac{nh}{l}\rceil$ ones, but it is difficult to extend this reasoning for counting the couples. Also, maybe using generating functions in some way like in this answer of mine.
I am especially interested in $n = \frac{l}{2}$ (or near to it) and $h \le \frac{l}{4}$ (about; for $h$ big enough it is clear that $c = 0$).
Any hint?
