Let $\textsf V = (\mathbb{Z}_2)^3$ and $F=\mathbb{Z}_2=\{0,1\}$.
How many different ordered basis does $\textsf V$ (over $F$) have? First of all, we know that the cardinality of $\textsf V$ is $2^3 = 8$ with
$$\textsf V = (\Bbb Z_2)^3 = \left\{ \begin{pmatrix} 0\\0\\0 \end{pmatrix}, \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \begin{pmatrix} 0\\0\\1 \end{pmatrix}, \begin{pmatrix} 1\\1\\0 \end{pmatrix}, \begin{pmatrix} 1\\0\\1 \end{pmatrix}, \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \begin{pmatrix} 1\\1\\1 \end{pmatrix} \right\}$$
Since $\dim (\textsf V) = 3$, basis come in $3$ vectors. I have manually listed all possible permutation, eliminating those that are linearly dependent, and the zero vector.
I got $210 - 18 = 192$ different ordered basis (unfortunately, I don't know how to show it more systematically).
Kindly help me with this. Next : how many distinct two-dimensional subspaces (over $F$) does $\textsf V$ have?
I think there are three two-dimensional subspaces, but I don't know how to show it. I get stuck here. Please help!