This post is just an elaboration of my comment to user52817's answer on the hat check problem. The rencontres numbers $D_{n,m}$ enumerate the permutations of $n$ items with exactly $m$ fixed points. Their number can be computed by multiplying the number of ways of choosing which items are the fixed ones by the number of derangements of the remaining items, that is, by multiplying the number of ways of choosing which $m$ people get their own hats by the number of ways of assigning hats not their own to the remaining $n-m$ people,
$$
D_{n,m}=\binom{n}{m}D_{n-m,0}=\frac{n!}{m!\,(n-m)!}\sum_{k=0}^{n-m}(-1)^k\frac{(n-m)!}{k!}.
$$
A standard expression for the number of derangements has been used on the right. (A derivation is at the end of this post.) As a consequence, the probability that a random permutation of $n$ items has exactly $m$ fixed points is
$$
\frac{D_{n,m}}{n!}=\frac{1}{m!}\sum_{k=0}^{n-m}(-1)^k\frac{1}{k!}.
$$
Fixing $m$ and letting $n$ go to infinity we get
$$
\lim_{n\to\infty}\frac{D_{n,m}}{n!}=\mathcal{N}\frac{1}{m!},
$$
where we can think of $\mathcal{N}=\sum_{k=0}^\infty(-1)^k\frac{1}{k!}$ as a normalization constant for the probability distribution. You will recognize that $\mathcal{N}=\frac{1}{e}$, but we can pretend we don't know that yet. Probabilities for all possible numbers of fixed points must sum to $1$, which gives
$$
1=\mathcal{N}\sum_{m=0}^\infty\frac{1}{m!}.
$$
So in addition to the expression for $\mathcal{N}$ already given, $\mathcal{N}$ is also the reciprocal of $\sum_{m=0}^\infty\frac{1}{m!}$. The consistency of these two characterizations can be verified by multiplying out the two sums and using an identity for the alternating sum of binomial coefficients to show that the product is $1$. (A version of this calculation for finite $n$ is given at the end of this post.) In addition to its role as normalization constant, $\mathcal{N}$ is the asymptotic probability that a random permutation has no fixed points, and also the asymptotic probability that it has exactly one fixed point. It should be noted that the probability distribution we have obtained by taking the $n\to\infty$ limit is the Poisson distribution with $\lambda=1$.
Whether students will see rencontres numbers as a compelling enough problem to motivate the definition of a new mathematical constant is a good question. My feeling is that the Poisson distribution is the fundamental thing since it shows up over and over again, with the rencontres numbers being but one application.
One potential objection that seems to have been raised is that Poisson probability is almost always defined as a limit of Bernoulli probability and that taking this limit requires the use of (a generalization of) the limit that shows up in the interest rate application. Which, I guess makes the Poison interpretation somehow "not independent" of the interest rate formula. All of this is discussed in David E Speyer's answer and the comments below it.
I'm actually not sure why that's so bad. Two points I would make are (1) that the rencontres numbers application shows that limits not involving the interest rate formula also arrive at the Poisson distribution, and (2) that it is just as interesting or more so to show interconnections between ideas as it is to show completely independent derivations. If the Poisson distribution is too complicated for students at this level to derive from scratch, it is still useful to them to teach some of the bare facts about it, such as that the probability of exactly $k$ hits is proportional to $\frac{\lambda^k}{k!}$.
With regard to the last point, it may be that students at the introductory calculus level will see the Poisson distribution as intimidating because it involves a transcendental expression $e^\lambda$. A key point is that the ratio of any two Poisson probabilities is just a power of $\lambda$ times a rational number--so the kind of thing that is very familiar from algebra. The only role the transcendental expression plays is as an overall normalization constant that divides every probability equally and that cancels out if you take ratios of probabilities.
So my view is that the Poisson distribution is very down-to-earth motivation for introducing the transcendental function $e^\lambda$ and the transcendental number $e$, the latter being relevant to the very important special case $\lambda=1$. (The importance of $\lambda=1$ is that, for a Poisson process unfolding in continuous time or continuous space, the inverse of the hit rate provides a natural time or distance scale for the problem. Hence one of the bald facts that I think students ought to know is that in a Poisson process the probability of zero hits within the natural time scale for the problem is $\frac{1}{e}$. (And also for one hit.) This is the most compelling occurrence of the constant $e$ "in nature" that I know of.
Some details filled in: The formula for $D_{n,0}$, the number of derangements of $n$ items, can be derived by defining $C_S$ to be the set of permutations of $1,2,\ldots,n$ in which each element of the set $S$ is a fixed point, so that $\lvert C_S\rvert=(n-\lvert S\rvert)!$. By inclusion-exclusion we have
\begin{align}
D_{n,0} &=n!-\sum_i\lvert C_{\{i\}}\rvert+\sum_{i<j}\lvert C_{\{i,j\}}\rvert-\sum_{i<j<k}\lvert C_{\{i,j,k\}}\rvert+\ldots+(-1)^n\lvert C_{\{1,2,\ldots,n\}}\rvert\\
&=n!-n(n-1)!+\binom{n}{2}(n-2)!-\binom{n}{3}(n-3)!+\ldots+(-1)^n\\
&=\frac{n!}{0!}-\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+\ldots+(-1)^n\frac{n!}{n!}.
\end{align}
The probability that a random permutation of $n$ items is a derangement is therefore
$$
\frac{D_{n,0}}{n!}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots+(-1)^n\frac{1}{n!}.
$$
That the probabilities for all possible numbers of fixed points sum to $1$ must, of course, work out for finite $n$. Let's see how this goes:
\begin{align}
\sum_{m=0}^n\frac{D_{n,m}}{n!} &=
\sum_{m=0}^n\frac{1}{m!}\sum_{k=0}^{n-m}(-1)^k\frac{1}{k!}\\
&= \sum_{t=0}^n\sum_{m=0}^t\frac{1}{m!}(-1)^{t-m}\frac{1}{(t-m)!}\\
&= \sum_{t=0}^n\frac{1}{t!}\sum_{m=0}^t\binom{t}{m}(-1)^{t-m}\\
&= \sum_{t=0}^n\frac{1}{t!}\delta_{t,0}=1.
\end{align}
In the second line we have introduced a new summation index $t=m+k$; in the fourth line $\delta_{a,b}$ is the Kronecker delta; it arises from the evaluation of the alternating sum of binomial coefficients.
