Number of overlapping columns

Question

Consider an $m \times n$ matrix $A$ with $m<n$. It is well-known that number of ways of choosing $k$ columns out of $n$ from $A$ is $\binom{n}{k}$, where ($k<m<n$). What is the number of ways of choosing $k$ columns from $A$ with only one overlapping column? Similarly, two overlapping columns, and so on to $k-1$ overlapping columns. Any help would be appreciated.

Answer 1

I think the answer is $\lfloor \frac{n-m}{k-m}\rfloor$, where $m$ is the number of common columns, $k$ is the number of columns chosen and $n$ is the total number of columns.

Here is the argument: But for the $m$ common columns, the remaining $k-m$ columns of any two pairs have to be mutually disjoint. And to maximize such set of choosings, the $n-m$ columns have to be split into as many parts as possible, with each part size $k-m$. This split will have size $\lfloor \frac{n-m}{k-m}\rfloor$.