# Five boys and five girls are to sit around a table. Find in how many ways this can be done if the boys and girls alternate

I have a query about this question:

Five boys and five girls are to sit around a table. Find in how many ways this can be done if the boys and girls alternate.

I know this question has been asked before, but I really don't understand why my specific solution is incorrect.

I thought that the answer would be $$4!5!*2$$, because you take the case that the initial person being seated is a boy, which has $$4!5!$$ permutations, then you multiply by $$2$$ to take into account the permutations, for which a girl is initially seated.

• I'm also confused to the question: The letters A, E, I, P, Q, R are arranged around a circle. Find the number of ways A is opposite R. I think the answer is (4!2!)/5!, however, the answers are 4!/5!. I have the same problem. Someone please help. Sep 26, 2021 at 11:58
• Sep 26, 2021 at 12:03
• i said i know that the question exists but I am not sure with my respective way of doing the question, can you clarify why my way of doing it is wrong? Sep 26, 2021 at 12:16
• The question is too unclear for me. For example, it does not mention whether it matter if a specific persion occupies a specific seat or not. Just "sitting at a round table" does not mean anything mathematically in that regard, and depending on interpretation you'll get different outcomes. Sep 26, 2021 at 12:23
• ...in your specific case it appears to me you are taking into account girls' permutation twice. You have 5 seats left thus $5!$ and not $2\cdot 5!$. Sep 26, 2021 at 12:25

## 1 Answer

Starting with boys or girls actually does not matter , because we only interested in the final solution ,i.e the number of distinct arrangements.However , you deal with how many ways there are to reach the same result. When you first place to girls (by $$(5-1)!$$) and after placing the boys by $$5!$$ , you obtain some arrangements whose result is equal to $$4! \times 5!$$ .Moreover, you obtain these exactly same arrangements when you place firstly boys and after the girls. The key point is that we are calculating the number of different arrangements here , not the number of approaches to reach them.Hence , putting an extra $$2$$ will cause count the same arrangements twice.