How to count number of sets of $d$ linearly independent vectors of length $n$ from $F_2^n$. I've been trying for quite a while to find out formula to get this number but I always seem to miss something.
Could you give me a hint in the right direction?
 A: Think about how to specify a sequence of $d$ linearly independent vectors in $F_2^n$.
Any nonzero vector $v_1$ will do for the first.  There are $2^n - 1$ such.
Any vector that is not in the span of $v_1$ will do for the second.  There are $2^n - 2$ of these, since the $F_2$ space spanned by $v_1$ has 2 elements.
Any vector that is not in the span of $\{v_1, v_2\}$ will do for the third.  The $F_2$ space spanned by $v_1$ and $v_2$ contains 4 elements.
Continuing in this way, we see that there are 
$$
(2^n - 1)(2^n-2)(2^n-4)\cdots(2^n-2^{d-1})
$$
sequences of length $d$ consisting of linearly independent vectors in $F_2^n$.  Each set with $d$ elements corresponds to $d!$ of these sequences.
A: Assuming $F_2 \sim \Bbb Z/2\Bbb Z$ :
Since for any vector $x\in F_2^n$, $x+x = 0$, for a collection of d linearily independant vectors $x_1, ..., x_d$, you have $2^d$ distinct vectors in $vect(x_1, ..., x_d)$.
Thus constructing a sequence of $d$ vectors means selecting one ($x_1$) among $2^n-1 = card(F_2^*)$, then one ($x_2$) among  $2^n-2^1 = card(F_2)-card(vect(x_1))$ and so on ... You then have all different sequence of $d$ linearly independant vectors where order matters. 
However each set will be counted $d!$ since order does not matter in sets. in the end you have :
$$\prod_{i=1}^d{2^n-2^i}\over d!$$
different sets
