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

The correct option is D $60480$

We want the $A$ and $B$ delegates to be seated next to each other while ensuring that $C$ and $D$ delegate aren't sitting together. This problem is difficult to solve directly so let's take an indirect approach.

First we count all possible arrangements where the $A$ and $B$ delegates are sitting together and subtract from them, those arrangements where $A$ and $B$ delegates are sitting together as well as $C$ and $D$ delegates are sitting together.

Let us consider $A$ and $B$ delegate as one unit this will ensure that they are always seated together. So now we have $9-1=8$ delegates. These can be seated in $8!$ ways. But the $B$ and $A$ delegate can interchange the seats among themselves so we have,

$2\times 8!=80640$ arrangements

By similar logic if we consider $A$ and $B$ as one unit and $D$ and $C$ as another unit we have $9-2-7$ delegates. So, 8 ! ways. But since they can interchange places we have,

$2\times 2\times 7!=20160$ arrangements

So, the number of arrangements where the $A$ and $B$ delegates are seated next to each other while the $C$ and $D$ delegate aren't sitting together are,

$80640-20160=60480$ arrangements