Enumerate all two-dimensional subspaces of the space $(\mathbb Z/2\mathbb{Z})^3$.
Obviously we have $|(\mathbb Z/2\mathbb{Z})^3|=2^3=8$ with
$(\mathbb Z/2\mathbb{Z})^3=\{v_0,\ldots,v_7\}=\left\{ \begin{pmatrix}0\\0\\0\end{pmatrix}, \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\right\}.$
However in the solution to the question is the assumption that there are seven different subspaces $\mathcal{C}_k\subset(\mathbb{Z}/2\mathbb{Z})^3,k\in\mathbb N,$ with $\dim(\mathcal{C_k})=2$:
$ \mathcal C_1:=\operatorname{span}\{v_1,v_2\}=\operatorname{span}\{v_1,v_3\}=\operatorname{span}\{v_2,v_3\}\\ \mathcal C_2:=\operatorname{span}\{v_1,v_4\}=\operatorname{span}\{v_1,v_5\}=\operatorname{span}\{v_4,v_5\}\\ \mathcal C_3:=\operatorname{span}\{v_1,v_6\}=\operatorname{span}\{v_1,v_7\}=\operatorname{span}\{v_6,v_7\}\\ \mathcal C_4:=\operatorname{span}\{v_2,v_4\}=\operatorname{span}\{v_2,v_6\}=\operatorname{span}\{v_4,v_6\}\\ \mathcal C_5:=\operatorname{span}\{v_2,v_5\}=\operatorname{span}\{v_2,v_7\}=\operatorname{span}\{v_5,v_7\}\\ \mathcal C_6:=\operatorname{span}\{v_3,v_4\}=\operatorname{span}\{v_3,v_7\}=\operatorname{span}\{v_4,v_7\}\\ \mathcal C_7:=\operatorname{span}\{v_3,v_5\}=\operatorname{span}\{v_3,v_6\}=\operatorname{span}\{v_5,v_6\} $
I did recognize the pattern that $\operatorname{span}\{v_a,v_b\}=\operatorname{span}\{v_a,v_c\}=\operatorname{span}\{v_b,v_c\}$ and I understand why this is true, but how do I get both the number of all subspaces AND the subspaces, too? My first thought was that the nullspace can be ignored if one wants to span spaces with two vectors - is that the reason why we have exactly seven subspaces?