Question

In how many ways can 9 persons sit around a table so that all shall not have the same neighbors in any two arrangements?

Answer 1

Hint: The total number of distinct ways in which n objects can be arranged in a circle is equal to $ (n-1)! $ .
There are 2 arrangements for every sequence of objects, one is clockwise and the other is anti-clockwise. Both these arrangements have the same neighbors for each object, because the sequence is the same.
The final answer would be $ \dfrac{(n-1)!}{2} $ .

Complete step-by-step answer:
This is a case of circular arrangement.
The total number of distinct ways in which 9 people can be arranged around a table will be equal to $ (9-1)!=8! $ .
Half of these are clockwise arrangements and the other half are anti-clockwise arrangements.
Any particular order of people will have the same set of neighbors in both the clock-wise and the anti-clockwise arrangements.
Therefore, the total number of ways in which no two arrangements have the same set of neighbors, will be: $ \dfrac{8!}{2} $ . Its value is equal to: $ \dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{2}=20160 $ .
So, the correct answer is “Option C”.

Note: The circular arrangement is also called a cyclic permutation.
 $ n! $ represents the product: $ n!=n\times (n-1)\times (n-2)\times ...\times 2\times 1 $
 $ 0!=1 $
Basic Principle of Counting:
If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possible ways for happening of events A and B are:
I.Either event A alone OR event B alone: $ m+n $ .
II.Both event A AND event B together: $ m\times n $ .