Question

In how many ways five people $A,B,C,D$ and $E$ be seated around a circular table if $A$ and $B$ must not sit next to each other?

• A. $12$
• B. $14$
• C. $16$
• D. $20$

Answer 1

The correct option is A $12$

$A$ $\xi$ $B$ not sit together $=$ Total $-\left(A\xi Bsittogether\right)$

$\Rightarrow$ Total permutation $=(5-1)!=4!=24$

$A$ $\xi$ $B$ sit together : $A\xi B$ should be considered as $1$ entity so now $4$ entities to be seated $=3!,$ But $A\xi B$ can interchange their places,

$\therefore 3!\times 2!=12$

$\Rightarrow$ Required $=24-12=12$

Hence, the answer is $12.$