I am working on the following problem for an applied probability qualifying exam.
Fix positive integers $m\leq n$ with $n>4$. Suppose $m$ people sit at a circular table with n seats, with all ${n \choose m}$ seatings equally likely. A seat is called isolated if it is occupied and both adjacent seats are vacant. Find the mean and variance of the number of isolated seats.
Letting $X$ be the number of isolated seats, I have written $X=\sum_{i=1}^{n}\mathbb{1}_{A_{i}}$, where $A_{i}=\{$$i^{th}$ seat is isolated$\}$ for i=2,...,n-1, and $A_{1}$ and $A_{n}$ defined appropriately given the circular arrangement. This yields $\mathbb{E}X=\sum_{i=1}^{n}\mathbb{P}(A_{i})$. Now I know that $A_{i}$ occurs only when chairs $i\pm1$ are vacant and chair $i$ is occupied.
I mistakenly initially computed that the probability is $p(1-p)^2$ where $p$ is the probability of a single chair being occupied (which turned out to be $m/n$). However, since these events are not independent, I am unsure how to compute the actual probability.
$A_i$ = 1 if $i$ th seat is isolated.
P($A_i$ = 1) = $\frac {n-3 \choose m-1}{{n \choose m}}$
Therefore E[ X ] = $ n \times \frac {n-3 \choose m-1}{{n \choose m}}$