Suppose I have $\ N $ people throwing their hats and then the hats are redistributed randomly. What are the chances no one gets back their original hat?
I try to start with $\ N = 3 $ and use this formula (because events are not disjoint). but I get $\ \frac{5}{27} $ which is not the answer.
$$\\ P(E_1\cup E_2 \cup E_3 \cup \dots \cup E_n) = \sum_{i=1}^n P(E_i) - \sum_{i_1<i_2}P(E_{i1} \cap E_{i2}) + \dots + \\ (-1)^{r+1} \sum_{i1<i2<...<ir} P(E_{i1} \cup E_{i2} \cup E_{i3} \cup ... \cup E_{ir}) + \dots + \\ (-1)^{n+1} P(E_1 \cap E_2 \cap E_3 \cap \dots \cap E_n ) $$
$$P(E_{i1}\cap E_{i2})=\frac1{n(n-1)}$$ because, once $i_1$ has the right hat, there are only $n-1$ hats that $i_2$ might get