Number of $n-1$-dimensional subspaces of $n$-dimensional space over finite field I got a question with two parts. 
Let $V$ be a $n$-dimensional vector space over $\mathbb{F}_{p}$ - finite field with $p$  elements.
a) How many $1$-dimensional subspaces $V$ has.
b) How many $n-1$-dimensional subspaces $V$ has.
I solved (a) with action of the multiplicative group of $\mathbb{F}_{p}$ on $V$, but I didn't succeed to solve (b) with similar idea..
I still prefer an idea with action of groups.. thanks !
 A: Hint: 
There is a bijection between subspaces of dimension $k$ and $k\times n$ matrices of rank $k$ in reduced row-echelon form.
A: This answer will only be helpful if you have some background in (projective) geometry.
The number of $1$-dimensional subspaces in a vector space of dimension $n$ over the field $\mathbb F_q$ corresponds to the number of points in a projective space of (geometric) dimension $n-1$ over the same field, denoted by $|\mathbb P^{n-1}(\mathbb F_q)|$. Since a projective space can be thought of as the disjoint union of an affine space of the same dimension and a hyperplane at infinity, i.e. a projective space of a dimension one smaller, we have that
$$ |\mathbb P^{n-1}(\mathbb F_q)| = |\mathbb A^{n-1}(\mathbb F_q)| + |\mathbb P^{n-2}(\mathbb F_q)| = q^{n-1} + |\mathbb P^{n-2}(\mathbb F_q)| . $$
Therefore, inductively, the answer is given by:
$$ |\mathbb P^{n-1}(\mathbb F_q)| = q^{n-1} + q^{n-2} + \dots + p + 1 = \frac{q^n-1}{q-1}. $$
Because of projective duality, a projective space has the same number of points and hyperplanes so this is actually the answer to the other question as well.
