Combinatorial/probabilistic meaning/analogy for $x^n / n!$

Question

The expression $x^n / n!$ appears in the infinite sum defining $e^x$ and similar terms in the sums defining $\cos(x)$, $\sin(x)$, etc. I would like to know if there is some combinatorial/probabilistic meaning or analogy to the term $x^n / n!$ and appropriately an example of a scenario of selection or decision making that could be represented by a consecutive sum of these terms similar to the infinite sums defining $e^x$ or $\cos(x)/\sin(x)$.

In other words: some probabalistic/combinatorical scenario whose calculation would converge to one of these functions ($e^x/\cos(x)/\sin(x)$).

Answer 1

If you divide every term by $e^{x}$ so the sum is $1$

then you get the Poisson distribution probability $e^{-x} \dfrac{x^n}{n!}$

of seeing $n$ items when the expected number of items is $x$.

Answer 2

If $X_1,X_2,\ldots,X_n$ are i.i.d $\text{Uniform}(0,1)$ random variables, then (check this for details)

$$P\left(\sum_{i=1}^n X_i \le x\right)=\frac{x^n}{n!}\quad,\, x\in [0,1]$$

In other words, $x^n/n!$ is the volume of the region $$\left\{(x_1,\ldots,x_n)\in \mathbb R^n: x_1,\ldots,x_n\ge 0, \sum_{i=1}^n x_i \le x\right\}$$ whenever $x\in [0,1]$. When $x=1$, this is related to the volume of a simplex.

We can also define a stopping rule associated with the $X_i$'s as

$$N=\min\left\{n: \sum_{i=1}^n X_i>x\right\} \quad,\, x\in [0,1]$$

Then, using the result above,

$$E(N)=\sum_{n=0}^\infty P(N>n)=\sum_{n=0}^\infty P\left(\sum_{i=1}^n X_i \le x\right)=e^x$$

This is in fact a popular exercise discussed here previously.