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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.

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  • $\begingroup$ 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. $\endgroup$
    – Blake
    Sep 26, 2021 at 11:58
  • $\begingroup$ What about math.stackexchange.com/questions/873676/… $\endgroup$
    – emacs drives me nuts
    Sep 26, 2021 at 12:03
  • $\begingroup$ 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? $\endgroup$
    – Blake
    Sep 26, 2021 at 12:16
  • $\begingroup$ 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. $\endgroup$
    – emacs drives me nuts
    Sep 26, 2021 at 12:23
  • $\begingroup$ ...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 $2\cdot 5!$. $\endgroup$
    – emacs drives me nuts
    Sep 26, 2021 at 12:25

1 Answer 1

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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.

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