Vector space over finite field Let's $\mathbb{F}_{p}^{5}$ is $5-$dimension space over $\mathbb{F}_{p}$, where $p$ is prime. How many ways can be decomposed the space $\mathbb{F}_{p}^{5}$ into a direct sum of two subspaces of dimension $2$ and $3$, i.e. present as
$$\mathbb{F}_{p}^{5}=V_1\oplus V_2,$$
$$\dim V_1 = 2,~~~\dim V_2 = 3.$$
Thanks.
 A: Let's do it in the, perhaps, most straightforward way. Any basis $e_1,\dots,e_5$ gives a desired decomposition: $V=\langle e_1,e_2\rangle\oplus\langle e_3,e_4,e_5\rangle$ -- but each decomposition can be obtained in multiple ways (which correspond to different choises of bases for $V_1$ and $V_2$). Namely, if $N_k$ is the number of bases in $k$-dimensional vector space, the answer is
$$
\frac{N_5}{N_2\cdot N_3}=
\frac{(q^5-1)(q^5-q)(q^5-q^2)(q^5-q^3)(q^5-q^4)}{(q^2-1)(q^2-q)\cdot (q^3-1)(q^3-q)(q^3-q^2)}=
q^6\frac{(q^5-1)(q^4-1)}{(q^2-1)(q-1)}
$$
(cf. Pierre-Yves Gaillard's comment; oh, and $q=p$, if you will).
So the answer is almost a q-binomial coefficient, but not quite: it's $q^6\binom 52_q$. This similarity can also be explained. Namely, consider a decomposition $V=V_1\oplus V_2$. A subspace $V_2'$ also gives a decomposition $V_1\oplus V_2'$ iff $V_2'\subset V_1\oplus V_2$ is transversal to $V_1$ -- i.e. $V_2'$ is a graph of some linear map $V_2\to V_1$. So for each 2-dimensional subspace there are exactly $|\operatorname{Mat}_{2\times 3}(F_q)|$ ways to complement it to a decomposition -- hence the answer $q^6\binom 52_q$.
A: What I've mean: Lat's $N_1(n, k)$ is the quantity of ordered sets consist of $k$ linear independent vectors (in $\mathbb{F}_{p}^{n}$). And let's $N_2(n, k)$ is the quantity of k-dimension sub-spaces of $\mathbb{F}_{p}^{n}$. It's obvious that
$$N_1(n, k) = (p^{n}-p^{0})(p^{n}-p^{1})(p^{n}-p^{2})\ldots(p^{n}-p^{k-1})$$
and
$$N_2(n, k)=\frac{N_1(n, k)}{N_1(k, k)}.$$
Because $N_1(k, k)$ is a quantity of different basis of $\mathbb{F}_{p}^{k}$.
As I see
$$N = N_2(5, 3)\frac{(p^{5}-p^{3})(p^{5}-p^{4})}{N_1(2,2)}.$$
