Probability of rolling every face of a die at least once in a sequence of N throws. Say you have a die with $k$ faces (each face has a probability $1/k$).
You throw this die $n$ times. 
What is the probability of having every single face showing at least once in that sequence?
 A: Imagine throwing sequentially. There are then $k^n$  equally likely outcomes. 
Now we count the favourables, in which each face appears at least once. This is a messy business, for which there is no closed form unless we use Stirling Numbers of the Second Kind (please see Wikipedia). Let $n\ge k$.
Let us count the bad outcomes, in which at least one face is missing. There are $(k-1)^n$ outcomes in which Face $1$ is missing, and the same number where Face $2$ is missing, for a total of $k(k-1)^n$. 
However, this double-counts the cases where two faces are missing. Which two faces are missing can be chosen in $\binom{k}{2}$ ways, and then the rest of the faces can be filled in in $(k-2)^n$ ways.
So our new estimate for the number of bad choices is $k(k-1)^n-\binom{k}{2}(k-2)^n$.
However, we have subtracted too much, for we have subtracted one too many times the outcomes where $3$ faces are missing. There are $(k-3)^n$ outcomes where three specific faces (at least) are missing. Add up over all the $\binom{k}{3}$ ways to choose three specific spaces.
 That gives a new estimate of $k(k-1)^n-\binom{k}{2}(k-2)^n+\binom{k}{3}(k-3)^n$.
Continue. 
A: The proof that the PIE argument yields the Stirling numbers times $k!$ i.e. that
$$\sum_{q=0}^k {k\choose q} (-1)^q (k-q)^N =
{N\brace k} \times k!$$
uses the integral
$$(k-q)^N = \frac{N!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{N+1}} \exp((k-q)z) \; dz$$
which gives for the sum
$$\frac{N!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{N+1}} 
\sum_{q=0}^k {k\choose q} (-1)^q \exp((k-q)z) \; dz
\\ = \frac{N!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{N+1}} (\exp(z)-1)^k \; dz.$$
This however is precisely $${N\brace k}\times k!$$
because the Stirling numbers of the second kind are the species
$$\mathfrak{P}(\mathcal{U}(\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the generating function
$$G(z,u) = \exp(u(\exp(z)-1)).$$
