(See also this similar [question](https://mathoverflow.net/q/434424/136218)).


Consider a $h \times 4n-h$ binary matrix (a matrix with all entries $a\_{ij}$, $1 \le i \le h$, $1 \le j \le 4n-h$, equal to $0$ or $1$). We know that each row has $2n$ entries equal to $1$ and $2n-h$ entries equal to $0$. All rows are different. Let $m = \lfloor log\_2{h}\rfloor$. There are a total of ${4n-h \choose m}$ possible sets of $m$ columns in it. Let $c(h, n)$ be the number of such sets of columns with indexes $1 \le j\_1 \lt j\_2 \lt \ldots \lt j\_m \le 4n-h$ such that $a\_{ij\_1} = 1 \lor a\_{ij\_2} = 1 \lor \cdots \lor a\_{ij\_m} = 1$, $1 \le i \le h$. It's like there is a path of ones from the top to the bottom of the matrix along the $m$ columns.


I would like to find a function $f(h,n)$ to have a good lower bound for $c(h,n)$:


$$c(h,n) \ge f(h,n)$$


Clearly $c(h,n) \ge 1$ because for each $r \le h$ rows there are $2nr/(4n-h) \gt r/2$ average ones per column, therefore there is a column with at least $r/2$ ones within those $r$ rows; starting with $r = r\_1 = h$, then $r = r\_2 = h/2$ and so on we can find $m$ columns satisfying the requirement.


Any hint other than trying to compute some values with some random examples?


A possible simplified version of the problem is fixing $h = n$.
Another case that I am interested in is $m=2$ instead of $m = \lfloor log\_2{h}\rfloor$.



Number of sets of columns "connecting" top to bottom of a matrix
Quite similarly to [my answer](https://mathoverflow.net/q/434465) to your other question, we have
$$c(h,n) \geq \binom{4n-h}m - h\binom{2n-h}m.$$
I'm not sure how good is this bound.


