This problem comes from my Master Thesis in combinatorial optimization.
Let $\pi \in \mathbf{S}_n$ be a permutation of $n$ elements and $r \in \{2,3,\dots,n\}$. Define the neihborhood $\mathbf{N}_r(\pi)$ centered at $\pi$ of radius $r$ as all the permutation that can be reached from $\pi$ exchanging $r$ indices, i.e.
$$ \mathbf{N}_r(\pi):= \Big\{\sigma \in S_n \, \bigg| \, \# \{i \, | \,\pi(i) \neq \sigma(i) \} = r \Big\} $$ I would like to prove that
$$\label{1} \tag{1} \#\mathbf{N}_r(\pi) = \binom{n}{r} r! \sum_{k=2}^r \frac{(-1)^k}{k!}, \qquad \text{for every $\pi \in \mathbf{S}_n$.}$$
I showed by hand that $$ \#\mathbf{N}_2(\pi) = \binom{n}{2} $$
Is there a theorical way to prove the result \ref{1}?
I think the problem can be transformed into some classic counting combinatorial problem.
As requested in the comments,
For any $r$, we build such a permutation by selecting the $r$ fixed points and then choosing a Derangement of the others. Thus the answer is $$\binom nr \,\times \,!(n-r)$$