In how many ways can a round table of $n$ knights impale each other? $n$ knights sit around a round table, which magically adjusts its size to fit the number of knights sitting around it. The knights cannot get up out of their seats and their swords are only long enough to impale their two immediate neighbours. Unfortunately due to an argument about Guinevere's honour, they have an enormous falling out and start impaling their immediate neighbours. Any impaling is immediately fatal and obviously a dead knight cannot impale either of his neighbours. Once a knight is dead, his seat vanishes and the table shrinks. At the end of the fight the table has shrunk to a single seat and all the knights but one are dead. Obviously there are $n$ possible outcomes, but how many different ways (in terms of who killed whom, and in what order) are there of arriving at any given outcome in which some given knight remains?

Suppose we fix one knight as the survivor from the outset.

When there are $k$ knights, with $k>2$, there are $k-1$ possibilites for the next dead, and 2 possibilities for his murderer. So there are $2(k-1)$ possible outcomes.

Total number of outcomes is actually the product $$\prod_{k=3}^n 2(k-1)=2^{n-2}(n-1)!$$