Halmos, Finite-Dimensional Vector Spaces, Sec. 7, Ex. 8 As a linear algebra refresher, I am working through the above cited text (2nd ed., 1958).  The exercise asks to determine the number of subsets of $\{0,1\}^3$ the are bases of $\mathbf{C}^3$ as a vector space over $\mathbf{C}$.  My approach was basically a somewhat smart enumeration:  List all the triplets (i.e., sets of three vectors) where a 1 appears as a component in each position in at least one vector, then eliminate those triplets which are linearly dependent.
Two questions:


*

*Is there a more elegant method?  Is there an efficient way to do this for any dimension?

*I counted 29.  Is this correct?

 A: First the set contains $8$ elements, of which $7$ can be used to form a basis (the set $(0,0,0)$ does not count).
So there are $\binom{7}{3} \, = 35$ the possible combinations.
A combination does not form a basis if one of the elements can be expressed as linear combination of the rest two of it. In this particular case, the only possibility is $v_1 = v_2 + v_3$ since the vectors are of component coefficients $0$ or $1$. Moreover, the two vectors $v_2$ and $v_3$ must be like $(0,0,1)$ and $(0,1,0)$ so that the sum does not have $2$ as component.
So there are merely $2$ possible situations. 


*

*$v_2$ and $v_3$ both have exactly one $1$ in their coordinates and in different places, like $(0,0,1)$ and $(0,1,0)$. There are $\binom{3}{2} = 3$ possibilities.

*$v_2$ has one $1$ and $v_3$ has two $1$s in their coordinates, and the $1$s are in a totally complementary position, for example $(0,0,1)$ and $(1,1,0)$, so that they add up to $(1,1,1)$. There are $\binom{3}{1} = 3$ possibilities.


So the answer is $35 - 3 - 3 = 29$.
A: Here's an attempt to invoke theory to solve this problem.
We have the following theorem.
Theorem. Let $V$ be an $n$-dimensional vector space over a field $\Bbb R$ with $p$ elements. Then the number of linearly independent subsets of $V$ consisting of $m$ elements is
$$
\frac{1}{m!}\prod_{k=0}^{m-1}\left(p^n-p^k\right)\Box
$$
Since $V=\{0,1\}^3$ is a $\Bbb Z/2$-vector space the theorem implies that the number of $\Bbb Z/2$-bases of $\{0,1\}^3$ is
$$
\frac{1}{3!}\prod_{k=0}^2\left(2^3-2^k\right)=\frac{1}{3!}\left(2^3-2^0\right)\left(2^3-2^1\right)\left(2^3-2^2\right)=28
$$
Since any subset of $\{0,1\}^3$ that is a $\Bbb Z/2$-basis of $\{0,1\}^3$ is also a $\Bbb C$-basis of $\Bbb C^3$ we have $28$ as a lower-bound for our desired number.
Now, the above theorem implies that the number of $3\times 3$ matrices whose columns are subsets of $\{0,1\}^3$ with odd determinant is $3!\cdot 28$. 
So, to finish our problem we need only count the number of $3\times 3$ matrices whose columns are subsets of $\{0,1\}^3$ with even determinant and divide by $3!$.
