Q. Delegates from $9$ countries including countries $A$, $B$, $C$ and $D$ are to be seated in a row. How many different sitting arrangements are possible if the delegates of countries $A$ and $B$ are to be seated next to each other and and delegates of countries $C$ and $D$ are not to be seated next to each other? How will the answer change if sitting is done at a round table?
I tried solving it as making a single entity of $AB$ and finding ways $= 8! \cdot 2!$, then subtracting the ways where $CD$ can sit together $= 7! \cdot 2! \cdot 2!$. Then my answer boils down to $8! \cdot 2! - 7! \cdot 2! \cdot 2!$, but answer book has $6! \cdot 7C2$ as answer. I am unable to find out where I went wrong.
You answer is correct. I am not sure what the book was thinking, but perhaps they meant to say $$ 6!\cdot 2!\cdot 7\cdot6. $$ The $6!\cdot 2!$ represents the ways to permute the seven delegates where $C$ and $D$ are excluded (treating $AB$ as a unit and using $2!$ to recover their orientation), while the $7$ is the number of ways to insert $C$ into this ordering, and the $6$ is the number of ways to insert $D$.