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Hi mathoverflow community, may be some one may give me a hint on the following problem before I spend much time on brute force search.

For $q$ a prime number and $n=6$, let $\mathbb {F}_{q}^{n}$ be an $n-$dimensional vector space over $GF_{q}$. Furthermore let $ U_i, U_j$ and $U_k$ be disjoint subspaces of $\mathbb{F}_{q}^n$ having each dimension $2$, with $i, j, k \in \{1,\dots, n\}$, $i\neq j \neq k$. What is the effective maximum number of disjoint subspaces of dimension $2$ in $\mathbb{F}_{q}^{n}$ so that $\langle U_i, U_j \rangle \cap U_k = \{0\}$?

Please note the keyword effective because using the Gaussian binomial coefficients it is possible to calculate the maximum number of all the subsets possible in $\mathbb{F}_{q}^{n}$ but probably not all fulfill the condition that $\langle U_i, U_j \rangle \cap U_k = \{0\}$.


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The spaces $U_1,\ldots,U_n$ are subspaces of $\mathbb{F}_q^n$ so any span of $U_i,U_j,U_k$ will again be a subspace of $\mathbb{F}_q^n$. What am I missing here? – Christian Stump Jul 3 '13 at 18:43
Also the next sentence: do you mean that we basically replace $n$ by $6$? Would it be possible to explain the situation better? – Christian Stump Jul 3 '13 at 18:46
Correct! Any span of $\{U_i, U_j, U_k\}$ is $\subseteq \mathbb{F}_{q}^{n}$, with $i, j, k \in \{1, \dots, n\}$. So The span of $\{U_i, U_j, U_k\}$ generate the subspaces $U_i, \dots, U_n$. And yes $n$ can be replaced by $6$. – R. Simeon Jul 3 '13 at 18:57
Please answer Christian's questions by editing the question to make sense. At the moment it doesn't. – Brendan McKay Jul 3 '13 at 21:46
Thanks for the feedback so far. I edited the question hopefully it makes more sense. – R. Simeon Jul 4 '13 at 10:09

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