Easy but nice. General statement wrote: Let $\mathcal{M}$ be the set of $n \times n$ real matrices of rank $m \le n$. Given a matrix in $\mathcal{M}$, the set of columns of $A$ has $2^n-1$ nonempty subsets. Let $k_A$ be the number of these subsets that are linearly independent. Then $ 2^m-1 \le k_A \le \binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{m}$ Lowerbound: Pick $m$ linear independent columns. All its non-empty subsets are linear independent, hence $k_A \ge 2^m-1$. Construction: \[ \left[ \begin{array}{c|c} I_{m \times m} & 0_{m \times (n-m) } \\ \hline 0_{(n-m) \times m} & 0_{(n-m) \times (n-m) } \\ \end{array} \right] \]Upperbound: Any subsets with more than $m$ columns are linear dependent, hence $k_A \le \binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{m}$. Construction: \[ \left[ \begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 3 & \cdots & n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 2^{m-1} & 3^{m-1} & \cdots & n^{m-1} \\ \hline 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{array} \right] \]