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I want to calculate how many bases of $\mathbb{C}^3$, as a complex vector space, there are in the subset of vectors whose coordinates are $0$ or $1$. I don't know how to approach this problem systematically, I tried some brute force method but obviously it gets too complex for large $n$, $\mathbb{C}^n$. Any hints?

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    You are considering $\mathbb{C}^3$ as a vector space over $\mathbb{C}$, or over $\mathbb{R}$ ( or some other filed)? – 2017-01-06
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    Sorry, I edited the post. – 2017-01-06
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    "It gets too complex for large $n$": pun intended? – 2017-01-06
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    According to [this MO post](http://mathoverflow.net/q/18636), the answer remains unknown for a general $n$. – 2017-01-06
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    @OpenBall Sure :-D – 2017-01-06
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    @user1551 Thanks for the link. – 2017-01-06
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    Using brute force search, we count 174 nonsingular $\{0,1\}$-matrices of size $3$. Therefore there are 174/6 = 29 *unordered* bases consisting of $\{0,1\}$-vectors in $\mathbb C^3$. – 2017-01-06
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    This problem is really easy when you are working over $\mathbb Z_2$ since you just want to count the total number of "bases", this trick lets you calculate the number of matrices with even determinant really easy, it is $7\times \times 4$. It turns out that there is an extra $3$ matrices with determinant equal to $2$ and an extra $3$ matrices with determinant $-2$. – 2017-01-06

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We can pick the first row in $7$ ways and the second row in $6$ ways so that the first two rows are linearly independant.

Now we just have to seperate into a couple of cases to check how many $1-0$ matrices are contained in the span of the first two rows.

The only way in which a fourth $1-0$ matrix is spanned is if the two rows do not share a $1$ or if one of the rows contains the ones of the other row.

So how many cases is this? First lets analyze the first case: In total there are $3^3$ ways this can happen, but there are $1+7+7$ cases in which the two rows are not linearly independent. So there are only $12$ cases in which the first two rows generate a fourth $1-0$ column.

For the second case first lets look at the cases in which the first row contains the second. In total there are $3^3$ cases, but not all of these satisfy that the two rows are linearly independent. We must subtract $1+7+7$ cases. So there are only $12$ cases to consider. We multiply this by $2$ to consider the case in which the second row contains the first one.

Hence the answer is $36\times 4+ (6\times 7 -36)\times 5=174$ ways.