Ways to choose a $ (n-k) $-dimensional linear subspace intersecting the given $ k $-dimensional subspace trivially 
Given an $ n $-dimensional linear space $ V $ and any set of basis of $ V $ say $ \{e_1,\cdots,e_n\} $, any partition of the basis elements into two subsets containing $ k $ and $ n-k $ elements determines appropriate subspaces $ P $ and $ Q $. We know that every $ k $-dimensional subspace $ W $ must have trivial intersection with $ Q $ for at least one of these partitions. My question is how many partitions at most can $ W $ intersect $ Q $ trivially.


To show there are at least one of these partitions, just note that if $ (v_1,\cdots, v_k) $ is linearly dependent $ k $-tuple in $ V $ with $ v_1\ne 0 $, then some $ v_i $ can be expressed as a linear combination of the preceding vectors $ (v_1, v_2,\cdots, v_{n-1}) $. And apply this fact to the ordered $ (k+n) $-tuple $ (w_1,\cdots,w_k,v_1\cdots,v_n) $ and using the fact that $$ \operatorname{Span}\{w_1,\cdots,w_k,v_1\cdots,v_n\}=V $$
and we can delete some $ v_i $ if it is a linear combination of its preceding vectors without changing the spanning space. At some point we get the desired $ n-k $ vectors of basis. 
I am wondering if there is a formula to compute the maximum number of ways of partitions so that $ W $ intersects $ Q $ trivially?
Edit:
I came up with this question when I was trying to figure out the Grassmannian is Hausdorff.
 A: In characteristic zero (at least), the answer is $\binom{n}{k}$ (all partitions).
Consider a matrix $A=(a_{ij})\in M_{k,n-k}(\mathbb{K})$ such that any minor is invertible. If such a matrix exists, then you can have a $W$ that intersects all $Q$ trivially for all partitions.
Indeed, for $i$ between $1$ and $k$, take $x_i=e_{n-k+i}+\sum_{j=1}^{n-k}a_{ij}e_j$, and $W$ spanned by the $x_i$.
Suppose we have a linear combination $\sum_{i\in I}\lambda_ix_i\in Q=\mathrm{Span}_{j\in J}(e_j)$, with $I$ a subset of indices between $1$ and $k$, and $J$ a subset of indices between $1$ and $n$ of cardinal $|J|=n-k$, such that $\lambda_i\neq 0$ for all $i\in I$. Then obviously $n-k+i\in J$ for all $i\in I$, so $J'=J\cap\{1,\ldots,n-k\}$ is of cardinal at most $n-k-|I|$.
Now if you consider $x'_i = \sum_{1\leq j\leq n-k, j\notin J'}a_{ij}e_j$ (which are morally the $x_i$ mod $Q$), we have $\sum_{i\in I}\lambda_ix_i=0$. But this gives a minor of size $|I|$ of the matrix $A$ which is not invertible, which is a contradiction.
Now, such a matrix exists over $\mathbb{Z}$ by taking a submatrix of a Vandermonde (see Rank of square matrix $A$ with $a_{ij}=\lambda_j^{p_i}$, where $p_i$ is an increasing sequence), so the argument works over any field of characteristic zero.
