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^{k-1}) 
\end{align*}
with $\binom{n}{k}_{q}=\frac{(q^{n}-1)\cdots(q^{n}-q^{k-1})}{(q^{k}-1)\cdots(q^{k}-q^{k-1})}$. 
My question is now, if it is possible to get a formula in the same fashion for the following number
\begin{align*}
\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*}.