3 red, 3 blue and 3 green beads are arranged in a circle. What is the probability that each bead has at least one neighboring bead of other color? All possible arrangements of the beads in the circle are 9!.
If we start from a specific bead to which we assign number 1 (starting point), then the number of possible arrangements of the 3 different colors is 
$\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{9!}{3!3!3!}$
It seems to me that it is impossible for ALL beads to have neighboring beads of the same color - am I missing something?
 A: Not all red beads have a neighbor of other color if and only if the red beads are placed on consecutive spots.

Let $R$ denote the event that the red beads are placed on consecutive spots.
Let $B$ denote the event that the blue beads are placed on consecutive spots.
Let $G$ denote the event that the green beads are placed on consecutive spots.
Then to be found is $1-\mathsf P(R\cup B\cup G)$ and with inclusion/exclusion and symmetry we find that this equals:
$$\begin{aligned}1-3\mathsf{P}(R)+3\mathsf{P}(R\cap B)-\mathsf{P}(R\cap B\cap G) & =1-\mathsf{P}\left(R\right)\left[3-3\mathsf{P}\left(B\mid R\right)+\mathsf{P}(B\cap G\mid R)\right]\\
 & =1-\frac{9}{\binom{9}{3}}\left[3-3\frac{4}{\binom{6}{3}}+\frac{2}{\binom{6}{3}}\right]\\
 & =\frac{41}{56}\\
 & \approx0,732142857
\end{aligned}
$$
Observe that there are $\binom93$ distinct triples of spots and by $9$ of them the spots are consecutive. 
This explains $\mathsf P(R)=9/\binom93$.
Under condition of event $R$ there are $6$ consecutive spots left so there are $\binom63$ triples in total of which $4$ are consecutive and $2$ of which are such that the events $G$ and $B$ will both occur ifthey are possessed by e.g. blue beads.
