Suppose there are n people invited to a party. Seats are assigned and a name card is made for each guest. However, floral arrangements on the table unexpectedly obscure the name cards. When the n guests arrive, they seat themselves randomly. Show that the probability that no one seats at the assigned seats, $P_n = \frac{n-1}{n} P_{n-1} + \frac1n P_{n-2}$.
My thoughts: $P_n = P(\text{n-1 people didn't sit at there assigned seats & the n-th person didn't sit at her assigned seat}) = P_{n-1} \times \frac{n-1}{n}$. But my thoughts are obviously incorrect. I'm just confused about why there would be a $P_{n-2}$ term. Could someone give me some hints? Thank you
Instead of considering full table try to consider emty table. We have two cases. First guest seats on place numbered $i$, next comes guest numbered $i$. He can seat on first place or other. If seats on place numered $1$. Then we have the problem reduced to $n-2@ places
In second case guest neither seats on $i$ and $1$. Still $n-1$ guests and $n-1$ places to seat.