Five boys and five girls are seated around a circular table, if the boys and girls are seated alternatively, then the number of arrangements is?

Question

Five boys and five girls are seated around a circular table, if the boys and girls are seated alternatively, then the number of arrangements is?

I tried doing the 'box method' in calculating the number of possibilities for this scenario, and I figured out the answer was $ 5! * 5! $ , although I think I'm wrong as the solutions say $ 5!/5 * 5! $ . Could someone please explain an efficient manner in completing this question?

Answer 1

see my answer: assume that the clockwise & anticlockwise arrangements are distinct.

first arrange all five girls or boys in a circle say girls are arranged in a circle such that a gap or a seat is vacant between two adjacent girls to seat a boy between them then the total number of circular arrangements of 5 girls $$=\frac{^{5}P_5}{5}=\frac{5!}{5}$$ now, the ways of seating boys in the five gaps of girls is just similar to arranging five boys in a line $=5!$

since both the events of arrangement are independent hence to tal number of arrangements $$=\frac{5!}{5}\cdot 5!$$

Answer 2

If we imagine the seats numbered from $1$ through $10$ around the table, so that they can each be individually identified even if we rotate the whole table, your answer is almost correct. The girls can be seated either in the even-numbered seats or in the odd-numbered seats ($2$ possibilities); in either case they can be seated in $5!$ different orders; and the boys can then be assigned to the remaining $5$ seats in $5!$ different ways, so we get a total of $2\cdot5!\cdot5!$ possibilities.

However, in problems about seating people around a circular table you’re generally supposed to assume that the seats are not individually identifiable, so that two seatings that arrange the participants in the same cyclic order are considered the same seating. A seating that puts Anna in seat $1$ is considered the same as a seating that puts her in seat $2,3$, or indeed any other seat, provided that each of the $10$ people still has the same left and right neighbors. Thus, each single seating in this sense corresponds to $10$ seatings if the seats are numbered, and we therefore have to divide by $10$ to get the correct number:

$$\frac{2\cdot5!\cdot5!}{10}=\frac{5!\cdot5!}5=4!\cdot5!\;.$$

This result can be obtained in another way as well. There are $4!$ different cyclic listings of the $5$ girls (because listings $ABCDE$ and $DEABC$ are the same cyclically), so we pick one of them and seat the girls in alternate seats around the table in that order. There are then $5!$ ways to assign the boys to the $5$ remaining seats, for a total of $4!\cdot5!$ possible arrangements. The assumption that only the cyclic order matters, not the assignment to specific seats, is taken care of at the first step in this approach.