Gaussian binomial coefficients, lattice paths, and vector spaces The Gaussian binomial coefficient ${n+k \choose k}_q$ gives a probability generating function for the number of lattice paths from $(0,0)$ to $(n,k)$ enclosing an area $a$ in the upper-right quadrant i.e. this count is given by the coefficient of $q^a$ in the corresponding series expansion of ${n+k \choose k}_q$.
For example, the number of lattice paths in the $10 \times 10$ box between the bottom left and top right corners is ${20 \choose 10}$, and those which enclose an area of 8 is the coefficient of $q^8$ in the corresponding series $${20 \choose 10}_q = 1 + q + 2 q^2 + 3 q^3 + 5 q^4 + \dots +2 q^{98} + q^{99} + q^{100}$$
which happens to be 22.
The Gaussian coefficient ${n+k \choose k}_q$ also counts the number of $k$-dimensional vector subspaces of an $n+k$-dimensional vector space over $F_q$.
What is the relation here? Is the vector space interpretation also somehow a generalisation of the notion of partitioning something under a set of constraints (here, into $k$ parts no larger than $n$)?
What, for example, does the lattice path generating function 'mean' when we set $q \neq 1$?
 A: I will illustrate a bijection from 


*

*$k$-dimensional subspaces of $F_q^n$.

*Monotone paths in an $(n-k)\times k$ box, where the cells above the path are each labeled with an element of $F_q$. [I hope it is apparent that these are counted by $\binom{n}k_q$.]


Every $k$-dimensional subspace of $F_q^n$ is the kernel of an $n\times n$ matrix with entries in $F_q$. Two matrices represent the same subspace iff they have the same reduced row echelon form (rref for short). Therefore, instead of counting subspaces, we may count rref matrices with exactly $n-k$ leading ones. 
In an rref matrix of rank $n-k$, the leading ones will occur in the first $n-k$ rows. In each row, the entries to the right of each leading one, and not above any other leading one, are arbitrary elements of $F_q$. For example, when $n=5$ and $k=2$, an example for how the matrix might look is
$$
\begin{bmatrix}
1&*&0&0&*
\\
0&0&1&0&*
\\
0&0&0&1&*
\\
0&0&0&0&0
\\
0&0&0&0&0
\end{bmatrix}
$$
The $*$'s represent arbitrary elements of $F_q$. If we ignore the bottom $k$ rows of the matrix (which are always zero), and ignore all columns containing leading ones, the result is
$$
\begin{bmatrix}
*&*\\
0&*\\
0&*
\end{bmatrix}
$$
Note that the boundary between the $0$'s and $*$'s is exactly a path in an $(n-k)\times k$ box where the cells above are labeled with entries of $F_q$. This is the bijection! I leave it to you to find the reverse map; basically, given a path with the cells above labeled with elements of $F_q$, you add a $1$ before each row of labels with a column of zeroes beneath.
Another example; when $n=2$ and $k=1$, a one dimensional subspace of $F_q^2$ is either the kernel of 
$$
\begin{bmatrix}
1&*\\
0&0
\end{bmatrix}\qquad \text{or}\qquad
\begin{bmatrix}
0&1\\
0&0
\end{bmatrix}
$$
There are $q$ matrices in the form on the left,  and $1$ matrix in the form on the right, for $1+q$ matrices, which is indeed $\binom{2}1_q$. The left matrix corresponds to the path in a $1\times 1$ box with an area of $1$, and the right corresponds to the path with an area of zero.
