When $q$ is a prime power, then on the one hand the [$q$-binomial coefficient](https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient) $\binom{n}{k}\_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}\_q^n$, and on the other hand it is the generating function of the sequence which sends $r$ to the number of words in two letters $X,Y$ of length $n$ with $k$ occurances of $X$ and with $r$ inversions (i.e. places where $Y$ comes before $X$). Therefore, there must a bijection


$$\{k\text{-dimensional subspaces of } \mathbb{F}\_q^n\} \\ \downarrow{\small\cong}\\ \coprod\_{r=0}^{k(n-k)} q^r \cdot \{\text{ words in X,Y of length } n \text{ with } k \text{ X's and } r \text{ inversions}\}$$


My question is if you can write down a nice, explicit bijection. I require that its description is independent of the theory of $q$-binomial coefficients, thus doesn't use the calculation of the cardinalities, and that it is also without recursion (since you can easily transform the inductive proof of the equality of cardinalities into a recursive bijection; this doesn't count here).


For $k=1$ the bijection looks as follows: It sends $\langle a\_1,\dotsc,a\_n \rangle$ to $(a\_1/a\_{r+1},\dotsc,a\_r/a\_{r+1}) \cdot Y^r X Y^{n-r-1}$, where $r$ is maximal with $a\_{r+1} \neq 0$.



Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions
You may act similarly as follows.


Let $V$ be a $k$-dimensional subspace of $\mathbb F\_q^n$. Take any its base, put its elements into the rows of some matrix, and make it to the reduced row echelon form $B$ (which is unique). The rows of $B$ still form a base of $V$.


Let $r\_1$, $r\_2$, $\dots$, $r\_k$ be the indices of leading elements in rows $1,2,\dots,k$. Then you put into correspondence to $V$ the tuple of all elements of $B$ which are non-fixed (i.e., those not in the leading columns, and not to the left of leading elements), along with the word $Y^{r\_1-1}XY^{r\_2-r\_1-1}XY^{r\_3-r\_2-1}X\dots XY^{n-r\_k}$ **read backwards**. The freedom of the coefficients provides the exponent of $q$ being
$$
 (n-r\_1-(k-1))+(n-r\_2-(k-2))+\dots+(n-r\_k),
$$
and the number of inversions (in the reversed word) is clearly
$$
 (n-r\_k)+(n-r\_{k-1}-1)+(n-r\_{k-2}-2)+\dots+(n-r\_1-(k-1)),
$$
so they trivially coincide.


