Circular permutations anyone?

M

Maths Pro

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The question reads : Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are never adjacent.
I don't know why I always feel confident after manually calculating the arrangements. I took total delegates to be from 5 countries as a prototype viz: ABCDE(It gave me 120 ways). I arranged them normally in a straight line and eliminated those arrangements which had AE or EA adjacent including those having A or E at the ends as clockwise or anticlockwise would not be seen as a distinct arrangement. Now I took those arrangements which have neither DC nor CD adjacent to each other including at the both ends. I am getting 40 arrangements.

Answer is supposed to be 12 ways using complement method( i.e Total number of circular arrangements of 5 delegates - Total number of arrangements in which CD are always adjacent.) If I want to know the number of arrangements without complement method is there a way?
First I would like to solve prototype.
 
I can see an ambiguity here: are rotated arrangements (e.g. ABCDE and BCDEA) considered different or identical?
 
I think the first issue is that for [imath]n[/imath] people seated in a circle, the number of permutations is [imath](n-1)![/imath]. This is assuming there is no outside reference point. Imagine the five people showing up one at a time. The first person sits anywhere (1 way). The second person then has 4 options, the next 3, etc. So for 5 people it is 4!.
 
As I understand it, circular permutations do not distinguish the two arrangements in #4, but do distinguish direction. So you may be making a mistake here:
Maths Pro said:
I arranged them normally in a straight line and eliminated those arrangements which had AE or EA adjacent including those having A or E at the ends as clockwise or anticlockwise would not be seen as a distinct arrangement.
Click to expand...
What definition were you given? What were you taught about circular permutations?

mathworld.wolfram.com

Circular Permutation -- from Wolfram MathWorld

The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over) circle is P_n=(n-1)!. The number is (n-1)! instead of the usual factorial n! since all cyclic permutations of objects are equivalent because the circle can be rotated...
mathworld.wolfram.com

So for 5 people, ABCDE, with A and B not adjacent, we can place A arbitrarily, and the put B in one of 2 seats that are not adjacent. Then C, D, and E can be placed in any of 3, 2, 1 seats, making a total of 1*2*3*2*1 = 12.

Using complements, there are 4! = 24 circular permutations without constraint, and 2*3! = 12 keeping A and B together (in either order), so there are 24 - 12 = 12 that fit the constraint.
 
So they give you both options, without using different terms. I wonder how they expect you to know how to interpret any particular problem?

For this problem, it depends on how they define "seating arrangement". (By the way, a "round table discussion" is not necessarily literal! So we may know nothing about the actual seating arrangement, which may not involve a round table at all. It can be done, for example, over Zoom.)
 
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