Number of two dimensional sub spaces of a vector space over a finite field. Let {$e_1,e_2,e_3,e_4$} br a basis of $4$-dimensional vector space over a finite field with p elements. The number of $2$-dimensional subspaces of $V$ not containing $e_4$ and not contained in Span({$e_1,e_2,e_3$}) is 
1) $(1+p+p^2)(p^2-1)$
2) $(1+p+p^2)(p^2+1)$
3) $(1+p+p^2+p^3)$
4) $(1+p)^3$
Hint is needed to solve the problem. Thank you.
 A: The correct is 1). I will regard a 2-dimensional subspace as the row space generated by the rows of a rank 2, $2 \times 4$ matrix 
$\left( \begin{array}{cccc}
a & b & c & d \\
d & e & f & g \\ \end{array} \right)$. 
Since such subspace must not be contained in the span of $e_1,e_2,e_3$ one of the generators, say the first row, must end with a $1$. So our subspace is 
$\left( \begin{array}{cccc}
a & b & c & 1 \\
d & e & f & 0 \\ \end{array} \right)$. 
In what follow I will use row elementary operations to get a better base of a subspace without explaining it in details.
Case 1  $f \neq 0$. Then our subspace is $\left( \begin{array}{cccc}
a & b & 0 & 1 \\
d & e & 1 & 0 \\ \end{array} \right)$. A subspace of this form contains $e_4$ if and only if $a=b=0$. So this case give us $p^4 - p^2$ subspaces not contained in $e_1,e_2,e_3$ and not containing $e_4$.
Case 2 $f = 0$ and $e \neq 0$.  Then our subspace is $\left( \begin{array}{cccc}
a & 0 & c & 1 \\
d & 1 & 0 & 0 \\ \end{array} \right)$. As in case in the previous case such a subspace contain $e_4$ iff $a=c=0$. So we get $p^3 - p$ more.
Case 3 $f = e =  0$.  Then our subspace is $\left( \begin{array}{cccc}
0 & b & c & 1 \\
1 & 0 & 0 & 0 \\ \end{array} \right)$. This give us $p^2 - 1$.
It is not difficult to see that the subspaces comming from 1),2) and 3) are different. So the total number is $p^4 - p^2 + p^3 - p + p^2 - 1 = (p^2 -1)(1 + p + p^2)$.
Here is another solution that explain the factorization $(p^2 - 1)(1+p+p^2)$ :
Let $W$ be a subspace not containing $e_4$ and not contained in span$\{e_1,e_2,e_3 \}$.
Consider the projection $PW$ of $W$ to span$\{e_1,e_2,e_3\}$ along $e_4$. Since $e_4$ does not belong to $W$ it follows that $PW$ is a 2-dimensional subspace of span$\{e_1,e_2,e_3\}$. We can recover $W$ from $PW$ and a nonzero element $\lambda$ of the dual space $PW^*$. Since there are $p^2 + p + 1$ 2-dimensional subspaces of span$\{e_1,e_2,e_3\}$ and the number of elements of the dual space of 2-dimensional subspace is $p^2$ the number of 2-dimensional subspaces we are looking for is $(p^2 - 1)(1 + p + p^2)$. 
