If we have vector space $V$ with dimension $n$ then how many subspaces of $V$ with dimension $m<n$ are there?
In my opinion the answer should be the number of ways to choose $m$ linearly independent vectors out of $n$ linearly independent vectors, regardless of the order of picking. This number is the binomial coefficient $\binom{n}{m}$?
If the vector space $V$ has fiinte dimension, say $n$, over a field of characteristic $0$, then the answer is infinte. But if the vector space has dimension $n$ over a finite field $\mathbb{F}_q$, then there are finitely many subspaces of $V$ having dimension $m$, with $m<n$. In fact, there are precisely
$$\frac{(q^n-1)\cdots(q^n-q^{m-1})}{|GL_m(\mathbb{F}_q)|}$$
subspaces of dimension $m$ in $V$.
In the (2,1) case over the reals, the question is "how many lines in the plane pass through the origin?".
Consider $m=1$ and $n=2$. Note that each subspace of dimension $1$ is a line through the origin in the plane. It is obvious that there are infinitely many such lines, and therefore infinitely many subspaces of dimension $1$.
In a more general setting, if $U \subset V$ is a subspace of $V$, $E$ is a basis of $U$ and $F$ is a basis of $V$, it is not necessarily true that $E \subset F$. Therefore it is impossible to generate all subspaces of $V$ with dimension $m$ by starting with a particular basis for $V$ and pick $m$ vectors in this basis and span the $m$ vectors.
