Parts a and b are trivial so those are of no concern. My conclusion for part c is that a proper q analog is the q binomial coefficient, counting the number of subspaces of dimension k in a vector space of dimension n whereas dimension n is over a finite field of q elements. However, this just seems like something "that is" and not something I can think of how to prove..
Let $N$ be an $\mathbb F_q$-vector space with dimension $n$.
$\binom{n}{m}_q\binom{m}k_q$ enumerates pairs of subspaces $(M,K)$ so that $K\subseteq M\subseteq N$, $\dim M=m$ and $\dim K=k$.
On the other hand, in order to choose $(M,K)$, you could first choose $K$ in $\binom{n}k_q$ ways, and then count the number of ways to choose $M$ for that given $K$. Subspaces of $N$ containing $K$ are in bijection with subspaces of $N/K$ (fourth isomorphism theorem). Therefore, the number of subspaces of $N$ containing $K$ with dimension $m$ is equal to the number of subspaces of $N/K$ with dimension $m-k$, or $\binom{n-k}{m-k}_q$.
Putting this altogether, we see that the original identity still holds when you stick little $q$-subscripts on all of the binomial coefficients.