Question

Three boys and three girls are to be seated around a table in a circle. Among them the boy X does not want any girl neighbour and the girl Y does not want any boy neighbour. How many such arrangements are possible?

• A. $5$
• B. $6$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

$4$

As boy $X$ does not need any girl neighbour so he will have $2$ boys sitting next to him.

The $2$ boys can be arranged in $2!$ ways.

Similarly $Y$ will be surrounded by $2$ girls and these $2$ girls can be arranged in $2!$ ways.

So number of arrangements $=2!\times 2!=2\times 2=4$

Hence, the answer is $4$.