3 boys and 3 girls are to be seated around a | vedclass

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(B) Let the boys be $B_1, B_2, X$ and the girls be $G_1, G_2, Y$.Since $X$ does not want any girl as a neighbour,$X$ must be seated between two boys,$

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$4$

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(B) Let the boys be $B_1, B_2, X$ and the girls be $G_1, G_2, Y$.Since $X$ does not want any girl as a neighbour,$X$ must be seated between two boys,$B_1$ and $B_2$.Since $Y$ does not want any boy as a neighbour,$Y$ must be seated between two girls,$G_1$ and $G_2$.This forces the arrangement to be a block of boys $(B_1, X, B_2)$ and a block of girls $(G_1, Y, G_2)$.In a circular arrangement of these two blocks,the number of ways to arrange them is $(2-1)! = 1! = 1$ way.Within the boy block,$B_1$ and $B_2$ can be arranged in $2! = 2$ ways.Within the girl block,$G_1$ and $G_2$ can be arranged in $2! = 2$ ways.Therefore,the total number of arrangements is $1 	imes 2! 	imes 2! = 4$.

$3$ boys and $3$ girls are to be seated around a circular table. Among the boys,$X$ does not want any girl as a neighbour,and among the girls,$Y$ does not want any boy as a neighbour. How many such arrangements are possible?

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