Question

Delegates from 9 countries include countries A, B, C, D are to be seated in a row. The number of possible seating arrangements, when the delegates of the countries A and B are to be seated next to each other and the delegates of the countries C and D are not to be seated next to each other is:
(A) $ 10080 $
(B) $ 5040 $
(C) $ 3360 $
(D) $ 60480 $

Answer 1

Hint: Calculate the number of possibilities for the delegates form the countries A and B sitting together while the delegates from the countries C and D are sitting randomly. Then calculate the number of possibilities for the delegates form the countries A and B sitting together while the delegates from the countries C and D are sitting together. Then subtract the two to get the answer.

Complete step-by-step answer:
We have delegates from 9 countries. Thus, we have a total of 9 delegates.
Now, we have to find out the number of arrangements of the seating arrangement of all the delegates with a condition that the delegates of countries A and B are always together while the delegates of countries C and D are not together.
First, let us calculate the number of arrangements when the delegates of the countries A and B are together.
For that, consider the delegates from the countries A and B as one unit. So now, there are total 8 units.
We know that, the number of possible arrangement of $ n $ different objects is given by $ n! $
Therefore, the possible arrangement for the $ 8 $ units will be $ 8! $
But, the delegates of countries A and B can replace places with each other and still can be together. Therefore, the number of arrangements for A and B to sit together will be $ 2! = 2 $
Hence, the total number of arrangements for all the delegates to sit while the delegates from the countries A and B are sitting together is $ 2 \times 8! $ . . . (1)
Now, let us consider the number of arrangement of the delegates when the delegates of countries A and B are sitting together and the delegates of the countries C and D are sitting together.
For this, we will consider, AB as one unit and CD as one unit. So now, there are total seven units.
Therefore, the total number of arrangements for the delegates to sit will be $ 7! $
But since, A and B can interchange seats with each other, the number of arrangements they can sit is $ 2! = 2 $
In the same way, the number of arrangements C and D can interchange seats with each other is also $ 2 $
Therefore, the total number of arrangements when AB and CD are always together will be $ 2 \times 2 \times 7! $ . . . (2)
Now, we can observe that the total number of arrangements for A and B to sit together but C and D to not to sit together will be the difference between the number of arrangements where only A and B are sitting together to the number of arrangements when AB and CD, both are sitting together.
Therefore, from equation (1) and equation (2), we can conclude that the number of sitting arrangements for the delegates when the delegates of the countries A and B are sitting together but the delegates of the countries C and B are not sitting together will be
 $ = 2 \times 8! - 2 \times 2 \times 7! $
By taking the common terms out and using property, $ n! = n(n - 1)! $ , we can write
 $ = 2(8 \times 7! - 2 \times 7!) $
 $ = 2 \times 6 \times 7! $
Now, $ n! = n(n - 1)(n - 2)(n - 3)....3 \times 2 \times 1 $
Thus, we can write the above expression as
 $ = 2 \times 6 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 $
Simplifying it, we get
 $ = 60480 $
thus, the number of sitting arrangements for the delegates when the delegates of the countries A and B are sitting together but the delegates of the countries C and B are not sitting together will be $ 60480 $
Therefore, from the above explanation, the correct answer is, option (D) $ 60480 $
So, the correct answer is “Option D”.

Note: In this question it is extremely important to understand that you cannot directly calculate the sitting arrangement of A and B sitting together while C and D are not sitting together, as there will be many possibilities. For example, C is sitting with delegate 1 while D is not, then D can sit with any of the 8 delegates. Then C is sitting with delegate 2, then D can sit with any of the other 8 delegates. Therefore, in this case, it is always better to find the total possibilities of arrangements for C and D to sit randomly and then subtract the possibility of C and D to be sitting together to get the possible arrangement of C and D never sitting together. This is a very important concept.