Count the number of bases of the vector space $\mathbb{C}^3$. Problem: Consider the set of all those vectors in $\mathbb{C}^3$ each of whose coordinates is either $0$ or $1$; how many different bases does this set contain?
In general, if $B$ is the set of all bases vectors then, 
$$B=\{(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)\}.$$
There are $8(6\cdot8+7)=440$ possible $B$s that contain unique elements with coordinates $0$ and $1.$ Now there is are $6\cdot 8+7$ sets that contain the element $(0,0,0)$, which makes the set $B$ linearly dependent and thus we are left with $385$ sets. Beyond this, I am finding it difficult to compute the final answer. Any hint/suggestion will be much appreciated.  

The set contains 29 different bases out of 35 possible permutations (excluding the zero vector).
Here are the six elements of the set which do not form a basis:
$$
\begin{eqnarray}
(0, 0, 1), (0, 1, 0), (0, 1, 1) \\
(0, 0, 1), (1, 0, 0), (1, 0, 1) \\
(0, 0, 1), (1, 1, 0), (1, 1, 1) \\
(0, 1, 0), (1, 0, 0), (1, 1, 0) \\
(0, 1, 0), (1, 0, 1), (1, 1, 1) \\
(0, 1, 1), (1, 0, 0), (1, 1, 1) \\
\end{eqnarray}
$$
