Number of different bases 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?
 A: 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.
