 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 neighbor and the girl Y does not want any boy neighbor. How many such arrangements are possible?$(a){\text{ 6}} \\ (b){\text{ 4}} \\ (c){\text{ 8}} \\ (d){\text{ 2}} \\$

Hint – In this question make the arrangement of the boy X with the remaining 2 boys only as he doesn't want any girl neighbor. In the similar way the arrangement of the girl Y has to be done with the two remaining girls as she doesn’t want any boy neighbor.

It is given that there are 3 boys and 3 girls.

It is given that one of the boys is X and one of the girls is Y.

Now let the other two boys be (b1 and b2) and the other two girls are (g1 and g2).

Now it is given among the boys X does not want any girl neighbor.

So the possible neighbor of X is the remaining boys (b1 and b2) and the number of arrangements of these boys is = (2!).

Now it is also given among the girls Y does not want any boy neighbor.

So the possible neighbor of Y is the remaining girls (g1 and g2) and the number of arrangements of these girls is = (2!).

So the possible number of such arrangements are

$\left( {2! \times 2!} \right) = 2 \times 1 \times 2 \times 1 = 4$

So this is the required answer.

Hence option (B) is correct.

Note – The tricky part in this question is that the total possible arrangements are multiplied as $2! \times 2!$ and not added as $2! + 2!$ because both the conditions of girl Y sitting with girls only and boy X sitting with x only are to be fulfilled. It’s somewhat similar to the concept of $A \cap B$ in set theory.