It is not difficult to get a formula for the number of $n\times n$matrices $g\in\mathrm{Mat}_{n}(\mathbb{F}_{q})$ with $\mathrm{rank}(g)=k$. Namely we have got \begin{align*} \mid\{g\in\mathrm{Mat}_{n}(\mathbb{F}\_{q})\mid \mathrm{rank}(g)=k\}\mid = \binom{n}{k}\_{q}\cdot(q^{n}1)\cdots(q^{n}q^{k1}) \end{align*} with $\binom{n}{k}_{q}=\frac{(q^{n}1)\cdots(q^{n}q^{k1})}{(q^{k}1)\cdots(q^{k}q^{k1})}$. My question is now, if it is possible to get a formula in the same fashion for the following number \begin{align*} f(n,q,k):=\mid\{g\in\mathrm{Mat}_{2n}(\mathbb{F}\_{q})\mid g=\begin{pmatrix} A & B \\\ C & A^{t} \end{pmatrix}, B=B^{t}, C=C^{t} \\text{ and }\mathrm{rank}(g)=k\}\mid \end{align*}. My conjecture (based on computer data for small $n$ and $k$) is the following, but I have no idea for the proof. \begin{align*} f(2n,q,2k)=\binom{2n}{2k}\_{q}\cdot q^{k(k+1)}\displaystyle\prod_{i=1}^{k}(q^{2i1}1), \end{align*} \begin{align*} f(2n,q,2k+1)=\binom{2n}{2k+1}\_{q}\cdot q^{k(k1)}\displaystyle\prod_{i=1}^{k}(q^{2i1}1) \end{align*}

Just for the (virtual) glory of possible reputation points, here is the comment as an answer. The columns can be switched to put B and C on the main diagonal, and multiplication of the bottom rows by 1 should produce a symmetric matrix from one of the desired form. This should be a bijective rank preserving operation which takes the current problem to one of counting symmetric matrices of a given rank over the chosen field. I am guessing that this new version is handled in the literature. Gerhard "Ask Me About System Design" Paseman, 2012.11.07 

