Q:

Number of vector subspaces in $C_{2}$

I have the vector space $C_{2}^3$ from the field $\mathbb{K} := C_{2}$
I want to name all vector subspaces of $C_{2}^3$.
$C_{2}^3= \Bigg\lbrace  \left(\begin{array}{c}0\\0\\0\end{array}\right),\left(\begin{array}{c}1\\0\\0\end{array}\right),\left(\begin{array}{c}0\\1\\0\end{array}\right),\left(\begin{array}{c}0\\0\\1\end{array}\right),\left(\begin{array}{c}1\\1\\0\end{array}\right),\left(\begin{array}{c}1\\0\\1\end{array}\right),\left(\begin{array}{c}0\\1\\1\end{array}\right),\left(\begin{array}{c}1\\1\\1\end{array}\right)\Bigg\rbrace$
And the basis is $\Bigg\lbrace \left(\begin{array}{c}1\\0\\0\end{array}\right) ,\left(\begin{array}{c}0\\1\\0\end{array}\right),\left(\begin{array}{c}0\\0\\1\end{array}\right)\Bigg\rbrace$
First I started to check all possibilities - that may work for this example but it isn't nice and doesn't work for bigger fields.
I want to know how many vector subspaces there are for a any finite vector space.
After a bit of searching I found out that I may sum up the number of vector subspaces from every dimension.
For the $0$-dimension there is obviously only $\Bigg\lbrace \left(\begin{array}{c}0\\0\\0\end{array}\right) \Bigg\rbrace$
For the $1$-dimension every element of $C_{2}^3$ should be a vector subspace. That are 7 more.
The $3$-dimension should be easy but I'm not sure is it the basis
$\Bigg\lbrace \left(\begin{array}{c}1\\0\\0\end{array}\right),\left(\begin{array}{c}0\\1\\0\end{array}\right),\left(\begin{array}{c}0\\0\\1\end{array}\right) \Bigg\rbrace$?
My main problem is the second dimension. There are $7!$ possibilities?
Not all of them are vector subspaces - some of them are the same. 
For example $\Bigg\lbrace \left(\begin{array}{c}1\\0\\0\end{array}\right),\left(\begin{array}{c}0\\1\\0\end{array}\right)\Bigg\rbrace$ and  $\Bigg\lbrace \left(\begin{array}{c}1\\1\\0\end{array}\right),\left(\begin{array}{c}0\\1\\0\end{array}\right)\Bigg\rbrace$
That's plausible for me but
how can I find out in general if two vector sub spaces are the same?
And the main part of this question: How can I calculate the number of vector subspaces for a specific dimension? 

A:

More generally, if the ground field is a finite field with $q$ elements, we obtain the number of $k$-dimensional vector spaces in $\mathbb{F}_q^n$ as follows:
We adopt the following $q$-analogue notations:
    $$
[n]_q = \frac{1-q^n}{1-q} = 1+q+ \cdots + q^{n-1},
$$
    $$
[n]_q!=[1]_q[2]_q\cdots [n]_q,
$$
    $$
\binom n k_q = \frac{[n]_q!}{[k]_q![n-k]_q!}.
$$

Theorem [Number of Subspaces]
Let $k\leq n$. Denote by $s(n,k)$ be the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$. Then
      $$
s(n,k)=\binom n k_q.
$$

Proof
First, we form a set of $k$ linearly independent vectors in $\mathbb{F}_q^n$. This can be done in $(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})$
    ways. Among these, there are $|\mathrm{GL}_k(\mathbb{F}_q)|$ sets yielding the identical subspaces. Therefore,
    $$
s(n,k) = \frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{(q^k-1)(q^k-q)\cdots (q^k-q^{k-1})}=\binom n k_q.
$$

