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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.

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    @MathCosmo - Thanks for the kind words! There are already a couple of outstanding answers, one of them having been accepted, and I don't see how to add anything interesting to them. The key word is "Orbit-Stabilizer Theorem". You can google this phrase, and/or look at the thread https://math.stackexchange.com/q/637665/660.2018-08-03

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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 $q^6\binom 52_q$.

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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 $N = N_2(5, 3)\frac{(p^{5}-p^{3})(p^{5}-p^{4})}{N_1(2,2)}.$

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    @KannappanSampath, why do you think that would be a problem? Given any 2-dimensional subspace surely there are several choices for the 3-dimensional complementary subspace?2012-02-09