Let $P = \{x \in R^n | Ax \geq b \}$. Suppose that at a particular basic feasible solution, there are $k$ active constraints with $k>n$. Is it true that there exists exactly $C(k,n)$ bases that lead to this basic feasible solution?
where,
$C(k,n) = \binom{k}{n}$
NO, it is not true. There is a maximum of $\binom{k}{n}$ bases because there are that many ways to choose $n$ out of $k$ possible bases, but not all combinations have to have linearly independent, so there may be less than $\binom{k}{n}$ bases.
As a counterexample, consider the following 2-dimensional feasible region: $$ x_1≥1, x_2≥2 \\ -x_1-x_2≥-4 \\ -2x_1-2x_2≥-8 \\ x_1, x_2≥0 $$
With $k=4$ and $n=2$, there are 6 ways to choose 2 constraints from the 4 constraints. However, only 5 lead to a BFS. The last 2 constraints are not linearly independent, and will not lead to a BFS.