Question

Three boys and three girls are to be seated around a circular table. Among them, the boy $X$ does not want any girl neighbour and the girls $Y$ does not want any boy neighbour. Then the number of possible arrangements is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is A $4$

Let $B_1,B_2$ and $X$ be three boys and $G_1,G_2$ and $Y$ be three girls.

Boy $X$ will have neighbours as $B_1$ and $B_2$ and the girl $Y$ will have neighbours as $G_1$ and $G_2$.

$B_1$ and $B_2$ can be arranged in $2!$ ways.

Also, $G_1$ and $G_2$ can be arranged in $2!$ ways.

Hence, required number of permutations $=2!\times 2!=4$