Find the probability that no two among $A$, $B$, and $C$ are together when $12$ people are arranged in a circle There are $12$ people including A,B and C. They are arranged in a circle. Find the probability that no two among A, B and C are together.
I have solved problem where cases involving two person sitting adjacent to each other. But here there are $3$ people where no two among them sits adjacent, hence I am getting confused.

We will seat $A$, $B$, and $C$ after we have seated the others.
Suppose one of the other people is person $D$.  Seat her somewhere.  It does not matter where since only the relative order of the people matters in a circular permutation.  As we proceed clockwise around the the table from person $D$, the other eight people may be seated in $8!$ ways.  This creates nine spaces in which to seat persons $A$, $B$, and $C$, one to the left of each of the nine people already at the table.  To ensure that no two of $A$, $B$, and $C$ are adjacent, we choose three of these nine spaces. Relative to person $D$, $A$, $B$, and $C$ can be arranged in these spaces in $3!$ ways.  Hence, the number of seating arrangements of the $12$ people at a round table in which no two of $A$, $B$, and $C$ are adjacent is 
$$8!\binom{9}{3}3!$$
Relative to person $D$, the other eleven people may be seated in $11!$ orders as we proceed clockwise around the table.  Hence, the probability that no two of $A$, $B$, and $C$ are seated together is 
$$\frac{8!\binom{9}{3}3!}{11!}$$ 
