Question

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

• A. $6$
• B. $4$
• C. $8$
• D. $2$

Answer 1

The correct option is B $4$

Let $B_1,B_2$ and X be three boys and $G_1,G_2$, Y be three girls.
Since X does not want any girl neighbour, $B_1,B_2$ can be their neighbours.
Similarly, the girl Y does not want any boy neighbour, therefore $G_1,G_2$ are the only neighbours of Y.
Now, $B_1,B_2$ can arrange themselves in $2!$ ways and $G_1,G_2$ can also arrange themselves in $2!$ ways.
$\therefore$ Required number of permutations $=2!\times 2!=4$