Working in $F_q^n$. How many different ways do we have to choose $l$ vectors such that every subset of size $k$ of them is linearly independent.
(Assume n is large)

My Progress: For the first k vectors, just keep out of the subspace spanned so far, so for $1 \leq i \leq k$ we have $q^n-q^{i-1}$ ways to choose the $i^{th}$ vector. But the next ones are harder.

Edit: Although the assignment asks me to find the exact number, a good lower bound can be helpful.

Hint $ $ Work universally, i.e. consider the matrix entries as indeterminates $\:\!\rm a_{\:\!ij},b_{\:\!ij}.\,$ Adjoin them to $\,\Bbb Z\,$ to get the polynomial ring $\rm R = \mathbb Z[a_{\:\!ij},b_{\:\!ij}].\, $ In this polynomial ring $\rm R,$  compute the determinant of $\rm\, (1+A B) A = A (1+BA)\,$ then cancel the nonzero polynomial $\rm\, det(A)\, $ (valid by $\rm R$  a domain). $ $ Extend to non-square matrices by padding appropriately with $0$'s and $1$'s to get square matrices. Note that the proof is purely algebraic - it does not require any topological notions (e.g. density).

Alternatively we may employ Schur decomposition as follows
$$\rm\left[ \begin{array}{ccc}
1 & \rm A \\
\rm B & 1 \end{array} \right]\, =\, \left[ \begin{array}{ccc}
1 & \rm 0 \\
\rm B & 1 \end{array} \right]\ \left[ \begin{array}{ccc}
1 & \rm 0 \\
\rm 0 & \rm 1\!-\!BA \end{array} \right]\ \left[ \begin{array}{ccc}
1 & \rm A \\
\rm 0 & 1 \end{array} \right]\qquad$$
$$\rm\phantom{\left[ \begin{array}{ccc}
1 & \rm B \\
\rm A & 1 \end{array} \right]}\, =\, \left[ \begin{array}{ccc}
1 & \rm A \\
\rm 0 & 1 \end{array} \right]\ \left[ \begin{array}{ccc}
\rm 1\!-\!AB & \rm 0 \\
\rm 0 & \rm 1 \end{array} \right]\ \left[ \begin{array}{ccc}
1 & \rm 0 \\
\rm B & 1 \end{array} \right]\qquad$$

See this answer for more on universality of polynomial identities, universal cancellation (before evaluation) and closely relation topics, and see also this sci.math thread on 9 Nov 2007.
