Consider seven people seated about a round table. How many circular arrangements are possible if at least one of them will not have the same neighbours in any two arrangements?
For answer , I used $ 6! = 720 $.
Since for circular permutation formula is $ \frac{n!}{n}$
But how do I find the number of circular arrangements are possible if at least one of them will not have the same neighbours in any two arrangements?
In the book , its given $360$ is answer.
Best Answer
Note that another way to phrase the question is: How many circular arrangements of seven people are possible if two arrangements are considered equivalent if each person has the same two neighbors?
The requirement that no person has the same two neighbors means that we need to count the number of distinguishable arrangements up to rotation and reflection.
Suppose Arjun is one of the seven people. Relative to Arjun, there are $6!$ ways to arrange the other six people as we proceed clockwise around the table from Arjun. This is the number of distinguishable arrangements up to rotation. However, each person would have the same two neighbors if we had proceeded anti-clockwise around the table from Arjun and selected the same people in the same order. Since reflections preserve the same neighbors, we must divide your answer by $2$. The number of distinguishable arrangements of seven people at a circular table up to rotation and reflection is $$\frac{6!}{2} = 360$$
Observe that a person will only have two neighbors if $n \ge 3$. For $n \ge 3$, the number of circular arrangements in which each person has the same two neighbors is $$\frac{(n - 1)!}{2}$$ When $n \leq 2$, each circular arrangement is its own reflection, so if $1 \le n \le 2$, there are $(n - 1)!$ distinguishable circular arrangements.