n Items Into n Boxes. $n$ items labeled $1, 2,...,n$ are placed randomly into $n$ boxes with the same labels, one item per box. Let $N$ be the number of items placed into a box with the same label (ie. item $1$ is placed into $box 1$, item $2$ in $box 2$, ...).
Find the expectation and variance of $N$, and calculate $P(N = 0)$.
 A: Hint: Try inclusion/exclusion and note symmetry.
A: The exponential generating function (EGF) of derangements is
$$\exp(-z) \frac{1}{1-z}.$$
This is because the labeled species here is
$$\mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=4}(\mathcal{Z})
+ \cdots)$$
which gives
$$\exp\left(\frac{z^2}{2}+\frac{z^3}{3}
+\frac{z^4}{4}+\cdots\right).$$
Hence we have for the number of derangements
$$n! [z^n] \exp(-z) \frac{1}{1-z}
= n! \sum_{q=0}^n \frac{(-1)^q}{q!} \sim n! \frac{1}{e}.$$
Thus $\mathrm{P}[N=0] = 1/e.$ 
The generating function of permutations with fixed points marked is
$$\exp\left(uz - z +\log\frac{1}{1-z}\right)$$
which is
$$G(z, u) = \exp(uz-z)\frac{1}{1-z}$$
This is because the species here is
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z}) +
\mathfrak{C}_{=2}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=4}(\mathcal{Z})
+ \cdots)$$
which gives
$$\exp\left(uz + \frac{z^2}{2}+\frac{z^3}{3}
+\frac{z^4}{4}+\cdots\right).$$
For the expectation $\mathrm{E}[N]$ compute
$$\left.\frac{\partial}{\partial u} G(z, u)\right|_{u=1}
= \left.z \exp(uz-z)\frac{1}{1-z}\right|_{u=1}
= \frac{z}{1-z}.$$
Hence $\mathrm{E}[N] = 1.$ For $\mathrm{E}[N(N-1)]$ we get
$$\left.\frac{\partial}{\partial u}
\frac{\partial}{\partial u} G(z, u)\right|_{u=1}
= \left.z^2 \exp(uz-z)\frac{1}{1-z}\right|_{u=1}
= \frac{z^2}{1-z}.$$
Hence $\mathrm{E}[N(N-1)] = 1.$ Finally we get for the variance
$$\mathrm{Var}[N] = \mathrm{E}[N^2] - \mathrm{E}[N]^2
= \mathrm{E}[N(N-1)] + \mathrm{E}[N] - \mathrm{E}[N]^2
= 1.$$
