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.  

Knowing that none of x,y,z can be (0,0,0) there are only 7 choices for each.  Since they must be different you only have $\binom{7}{3}=35$ choices to make.
This is small enough to sort through by hand.
