Given the set of numbers $A = \left \{ 1, 2, 3, 4, 5, 6, 7, 8 \right \}$. How many permutations leave exactly four numbers fixed?
I've considered this as a solution... but then I stumbled upon a thought that I might count the same permutation multiple times.
$$8! - \binom{8}{4}\left | \bigcup_{i=1}^{4}A_{i} \right |$$ ?
Any thoughts?
With the purpose of getting a value to verify our work we obtain using combinatorial classes the following class $\mathcal{P}$ of permutations with fixed points marked:
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{P} = \textsc{SET}( \mathcal{U} \times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \cdots).$$
This gives the EGF
$$F(z, u) = \exp\left(uz+\frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \cdots \right) \\ = \exp\left(\log\frac{1}{1-z} + (u-1)z\right) = \frac{1}{1-z} \exp\left((u-1)z\right) \\ = \frac{1}{1-z} \exp\left(-z\right) \exp\left(uz\right).$$
Hence permutations with exactly $k$ fixed points have EGF
$$G(z) = \frac{z^k}{k!} \frac{\exp(-z)}{1-z}.$$
Here we have $k=4$ and we get with $n=8$
$$8! [z^8] \frac{z^4}{4!} \frac{\exp(-z)}{1-z} = 1680 [z^4] \frac{\exp(-z)}{1-z} \\ = 1680 \sum_{q=0}^4 [z^q] \exp(-z) [z^{4-q}] \frac{1}{1-z} = 1680 \sum_{q=0}^4 \frac{(-1)^q}{q!} \\ = 1680 \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 630.$$
This is
$${8\choose 4} D_4$$
where $D_n = n! \sum_{q=0}^n \frac{(-1)^q}{q!}$ counts derangements.