I just came across this permutation question that confused me.
The question: How many ways are there to arrange 5 girls and 5 boys around a circular table with 10 chairs, assuming: (symmetrical arrangements around the table are identical or in other words, the chairs are identical)
(a) No restrictions
(b) Boys and girls alternate
(a) So there are 10! ways to arrange the 5 boys and girls, assuming each chair is unique. But in this case, all the chairs are identical, so we divide by 10 to eliminate the identical symmetrical circular arrangments. $\frac{10!}{10}=362,880$
(b) This question is the one that confuses me. If the boys and girls have to alternate then the number of permutations would be $\underline{5}\times\underline{5}\times\underline{4}\times\underline{4}\times\underline{3}\times\underline{3}\times\underline{2}\times\underline{2}\times\underline{1}\times\underline{1}=5!\times5!=14400$. But in order to account for identical symmetrical circular arrangements, you would have to divide by 10 ---> $\frac{14400}{10} = 1440$.
But apparently the answer is actually 2880, which is double what I got. Can someone explain what was wrong in my logic? I can see if I did $\frac{14400}{5} = 2880$ it would be give the right answer, but I have no reason why.
ancientmathematician has explained the flaw in your reasoning in the comments.
Since we only care about the relative order of the people, we can avoid dividing by symmetry arguments by fixing a particular person and counting arrangements relative to her.
Suppose Angela is one of the girls. Seat her. Doing so determines which seats are occupied by girls and which seats are occupied by boys. The other four girls may be seated in $4!$ ways as we proceed clockwise around the table from Angela. The five boys may be seated in $5!$ ways as we proceed clockwise around the table from Angela. Hence, there are $4!5!$ distinguishable seating arrangements.