Question

The number of ways in which 7 men can sit in a round table such that every person shall not have the same neighbours in any two arrangements is:
A. 360
B. 720
C. 700
D. 300

Answer 1

Hint: The above question requires the understanding of circular permutations, it is the number of ways of arranging n persons at a round table i.e in a circle. A circular permutation can be performed in two ways given as clockwise and anticlockwise.

Complete step-by-step answer:
Since in this question we have to arrange persons in a circle and 7 persons have to be arranged in a circle so that every person shall not have the same neighbor.
We know that the seats are not numbered at a round table and there is no distinction between the first and the last seat, only changes in the relative positions are required. Therefore we fix one of the person out of seven at one position and then arrange the remaining (n-1) persons in $ (n - 1)! $ where the value of n is 7:
 $
   \Rightarrow (7 - 1)! = 6! \\
   \Rightarrow 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \;
 $

In the above circular permutation we have distinguished between clockwise and anti clockwise where every person has got the same neighbor but we do not need this we have been asked to arrange so that no person has the same neighbor which can be done as:
 $
  \dfrac{{(n - 1)!}}{2} = \dfrac{{6!}}{2} \\
   \Rightarrow \dfrac{{720}}{2} = 360 \;
 $
Hence there are $ 360 $ ways to do the above arrangement and therefore the correct option is A.
So, the correct answer is “Option A”.

Note: The two ways in circular permutation, clockwise and anticlockwise have only one distinction where the persons sitting have a right hand neighbor in one case that changes to the left hand neighbor in the second case for the other person.