What is the number of subspaces of dimension $k$ in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$? What is the number of subspaces of dimension $k$ in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$?
See What is the number of basis in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$?
It is $\frac{(p^n -1)(p^n - p)...(p^n - p^{k-1})}{(p^k -1)(p^n - p)...(p^k - p^{k-1})}$ with $(p^n -1)(p^n - p)...(p^n - p^{k-1})$ the number of $k$ independent vectors.
Then, how many subpaces are in direct sum with this dimension $k$ - subspace ?
I think it would be $\frac{(p^n -p^k)(p^n - p^{k+1})...(p^n - p^{n-1})}{(p^k -1)(p^n - p)...(p^k - p^{k-1})}$ for the same reasons but the initial choice of independent vectors is more restrictive.
Do you have an idea about a recurrent formula that represents the number of families $(e_1,...,e_n)$ of rank $r$, with $r≥0$ and $k≥1$ ?
There are several questions but all are linked (so it would be difficult to follow different posts). 
 A: The number of $k$-dimensional subspaces of $\mathbb{F}_{q}^{n}$ is given by the Quantum Binomial Coefficient ${n \brack k}_{q}$. This quantity has algebraic significance, as the stabilizer of a group action. For me, the Orbit-Stabilizer Theorem provides stronger justification as to why we divide by the number of ways to fix a $k$-dimensional subspace. 
Let's begin with some terminology.
Definition 1: The Quantum Factorial ($q$-factorial) is the function:
$$[n!]_{x} = \prod_{i=1}^{n} \left( \sum_{j=0}^{i} x^{j} \right) = \prod_{i=1}^{n} \frac{x^{i}-1}{x-1} $$
Definition 2: The Quantum Binomial Coefficient ($q$-binomial coefficient) is the function:
$${n \brack k}_{x} = \frac{[n!]_{x}}{[k!]_{x} \cdot [(n-k)!]_{x}}$$
The Quantum Factorial and Quantum Binomial Coefficient are closely related to the combinatorics of finite vector spaces, when evaluated at $q = p^{n}$ for a prime $p$. 
Claim 1: $\text{GL}_{n}(\mathbb{F}_{q}) = \prod_{i=0}^{n-1} (q^{n}-q^{i})$.
Proof: I will leave this as an exercise. [Hint: Count the number of ways to obtain $n$ linearly independent vectors in $\mathbb{F}_{q}^{n}$].
Remark: Observe that:
$$\prod_{i=0}^{n-1} (q^{n}-q^{i}) = \prod_{i=0}^{n-1} q^{i}(q^{n-i}-1) = q^{\binom{n}{2}} \prod_{i=1}^{n} (q^{i}-1) = q^{\binom{n}{2}} [n!]_{q} (q-1)^{n}$$
Claim 2: ${n \brack k}_{q}$ counts the number of $k$-dimensional subspaces of $\mathbb{F}_{q}^{n}$, which we denote $\text{Gr}_{n}(k, \mathbb{F}_{q})$. 
Proof: We let $\text{GL}_{n}(\mathbb{F}_{q})$ act on $\text{Gr}_{n}(k, \mathbb{F}_{q})$. We first show this action is transitive. Recall that any two $V, W \in \text{Gr}_{n}(k, \mathbb{F}_{q})$ are isomorphic. Let $T : V \to W$ be an isomorphism. We extend $T$ to an automorphism of $\mathbb{F}_{q}^{n}$ as follows. Let:
$$M = \begin{pmatrix} [T]_{k \times k} & B_{k \times (n-k)} \\ O & C_{(n-k) \times (n-k)} \end{pmatrix}$$
Where $[T]$ is the matrix representation of $T$, and we extend the remaining $n-k$ columns to complete an ordered basis of $\mathbb{F}_{q}^{n}$. So $\text{GL}_{n}(\mathbb{F}_{q})$ acts transitively on $\text{Gr}_{n}(k, \mathbb{F}_{q})$. 
Fix $V = \langle e_{1}, \ldots, e_{k} \rangle \in \text{Gr}_{n}(k, \mathbb{F}_{q})$, where $e_{1}, \ldots, e_{k}$ are the first $k$ standard basis vectors. By the Orbit-Stabilizer Lemma, we have that:
$$|\text{GL}_{n}(\mathbb{F}_{q})| = |\text{Stab}(V)| \cdot |\mathcal{O}(V) = \text{Gr}_{n}(k, \mathbb{F}_{q})|$$
If we can determine $|\text{Stab}(V)|$, we are done. Note that any element of $\text{GL}_{k}(\mathbb{F}_{q})$ fixes $V$. Let $T \in \text{GL}_{k}(\mathbb{F}_{q})$. We extend $T$ to an element of $\text{GL}_{n}(\mathbb{F}_{q})$ in the same way as described above. There are $|\text{GL}_{k}(\mathbb{F}_{q})|$ ways of selecting $T$, and $\prod_{i=k}^{n-1} (q^{n}-q^{i})$ ways of selecting the remaining $n-k$ columns. Thus:
$$|\text{Stab}(V)| = \left(\prod_{i=0}^{k-1} (q^{k}-q^{i}) \right) \cdot \left( \prod_{i=k}^{n-1} (q^{n}-q^{i} \right)$$
Which is equal to:
$$\left( [k!]_{q} (q-1)^{k} q^{\binom{k}{2}} \right) \cdot \left(q^{\binom{n}{2} - \binom{k}{2}} [(n-k)!]_{q} (q-1)^{n-k} \right)$$
This is equal to: $[k!]_{q} (q-1)^{n} [(n-k)!]_{q} q^{\binom{n}{2}}$. 
And so:
$$|\text{Gr}_{n}(k, \mathbb{F}_{q})| = \frac{[n!]_{q} q^{\binom{n}{2}} (q-1)^{n}}{[k!]_{q} (q-1)^{n} [(n-k)!]_{q} q^{\binom{n}{2}}} = {n \brack k}_{q}$$
QED. 
A couple comments that aren't immediately related to your question, but are good insight. First, as $q \to 1$, you recover the standard factorial and binomial coefficient from the $q$-factorial and $q$-binomial coefficient. So in a sense, the Symmetry group behaves as a special case of the General Linear group, if we had a field of one element. In an introductory algebra class, the Symmetry group appears to be the "biggest" and most important group, in large part due to Cayley's theorem. However, we are seeing that matrix groups are more general than permutation groups. In light of this, as well as the fact that linear algebra is very well understood, matrix groups are arguably the "right" objects to study groups. This provides a lot of motivation for linear representation theory.
