Combinatorics in finite vector space 
Let $q$ be a prime power and $V$ a finite $\mathbb F_q$-vector space with two subspaces $I$ and $J$.
  Let $k$, $a$ and $b$ be non-negative integers.
  Determine the number of subspaces $K$ of $V$ with $\dim(K) = k$, $\dim(K\cap I) = a$ and $\dim(K\cap J) = b$.

The answer should depend on the numbers $a,b,k$ and $v = \dim(V)$, $i = \dim(I)$, $j = \dim(J), c = \dim(I\cap J)$, but otherwise be independent of the exact choice of $I$ and $J$.
In the case that only a single subspace $I$ is given (say $J = \{0\}$ and $b = 0$), I found the answer
$$
q^{(k-a)(i-a)}\binom{i}{a}_{\!q} \binom{v-i}{k-a}_{\!q},
$$
where $\binom{x}{y}_q$ denotes the Gaussian binomial coefficient.
However for two subspaces $I$ and $J$, this problems looks quite hard to me. Any ideas are welcome. Even the special cases $I \cap J = \{0\}$ or $J \subseteq I$ (both generalizing the case $J = \{0\}$ discussed above) are not clear to me.
 A: It is usually easier to convert these kind of problems to finite projective geometry, and to look at your subspaces as projective spaces. There is an entire branch of research dedicated to finite projective geometry and I think these basic questions are all solved there. In what I write below, I have ignored the difference of $1$ between vector and projective dimension, to keep it readable for you (as I presume you are more used to vector geometry).
I don't have the books here, so I can't give you a ready formula, but the main idea is always the following: pick an arbitrary I,J with your properties. Now, denote $C=I\cap J$ and denote an extra variable $t=\dim(I\cap J\cap K)$. We will first compute the requested number for a given value of $t$. This is much easier (and that's probably why you succeeded on the $J=\{0\}$ case, as here $t$ is fixed).
Since we know the group actions, we know that indeed all choices are projectively equivalent, hence indeed the number will be independent of the exact spaces chosen. If you want you can even give them special coordinates of your choosing.
Then, within I and J, compute the number of possibilities for $K\cap I$ and $K\cap J$. The number of $t$-spaces in $I\cap J$ is $N_1(t)$, this can be computed. Now, fix any such $t$-space $T=I\cap J\cap K$. Each such choice will yield the same number, so it's merely taking a product with $N_1$ at the end. 
In $I$, one can compute the number of $a$-spaces $A$ which have $A\cap C=T$, denote this number by $N_2(t)$. In $J$, one can compute the number of $b$-spaces $B$ which have $B\cap C=T$, denote this number by $N_3(t)$. Now fix such subspaces $A$ and $B$ (again, it doesn't matter which one, since all choices within these constraints are projectively equivalent).
Now, all that is left is counting how many $k$-spaces $K$ there are with $A=I\cap K$ and $B=J\cap K$. This is equivalent to saying that $\langle A,B\rangle = \langle I,J\rangle \cap K$. Denote $X=\langle A,B\rangle$ and $Y=\langle I,J\rangle$, then the problem that we are left with is "Determine the number of $k$-spaces $K$ which intersect $Y$ in $X$." This number is fixed (since $X$ and $Y$ are fixed) and should be a lot easier than your original problem; denote it by $N_4(t)$.
Then the total number for given $t$ is $N_1(t)\cdot N_2(t)\cdot N_3(t)\cdot N_4(t)$, and the original total number you requested (which is for any $t$) is just $$\sum_t N_1(t)\cdot N_2(t)\cdot N_3(t)\cdot N_4(t),$$ where the summation is taken over all nonzero values of the product (there are only finitely many).
