Fixed-point-free permutations

The permuations of the elements of {1,2,3} are followed by the permutioncycle(s)
1-2-3 = (1)(2)(3)
1-3-2 = (1)(23)
2-1-3 = (12)(3)
2-3-1 = (123)
3-1-2 = (132)
3-2-1 = (13)(2).
Only two of them are so called "fixed-point-free", i.e. they have no cycles of length 1: 2-3-1 and 3-1-2. There is a symbol for this number D for derangement. So D(3)=2. We can easily find D(4) by thinking in permutation cycles: those that do not include length 1 cycles are fixed-point-free. The possible cycle structures of permutations of length 4 are 1-to-1 related to the partitions of 4 ( number of permutations ).
4 = xxxx ( 6 = 4*3*2*1 / 4 )
31= xxx-x ( 8 = 4*3*2 / 3 )
22 = xx-xx ( 3 = 4*3 / 2 * 2 )
211 = xx-x-x ( 6 = 4*3 /2)
1111 = x-x-x-x ( 1 = 1 )
All basic M208/GTA1 material, I suppose. This formula is however not in the course.
$$D(n) = n! \cdot \sum_{k=0}^{n} \frac{(-1)^k}{k!}$$
With the formula for D we can simply check the result for 4.
$$D(4) = 4! \cdot (
\frac{(-1)^0}{0!} +
\frac{(-1)^1}{1!} +
\frac{(-1)^2}{2!} +
\frac{(-1)^3}{3!} +
\frac{(-1)^4}{4!} ) =9.
$$