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. The number of such arrangements possible is

• A. $4$
• B. $6$
• C. $8$
• D. None of these

Answer 1

The correct option is A $4$

As shown in figure $1,2$ and $X$ are the three boys and $3,4$ and $Y$ are three girls, Boy $X$ will have neighbors as boys $1$ and $2$ and the girls $Y$ will have neighbors as girls $3$ and $4$.

$1$ and $2$ can be arranged in $P(2,2)$ ways $=2!=2\times 1=2$ ways.

Also, $3$ and $4$ can be arranged in $P(2,2)$ ways $=2!=2\times 1=2$

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