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Say there are n boys and n girls seated in a circle with boys and girls alternating. What will be the probability that no girl is sitting beside her brother and no boy is sitting beside his sister. Each boy has only one sister and each girl has only one brother.

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  • $\begingroup$ By the way, this is known as the ménage problem and is very well-studied. Go check Wikipedia! $\endgroup$ – Parcly Taxel Oct 18 '16 at 4:31
  • $\begingroup$ Use inclusion exclusion principle. $\endgroup$ – Majeed Siddiqui Oct 18 '16 at 4:31
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WRONG SOLUTION

The no of ways $2n$ boys and girls can seat in the circle=$(2n-1)$!


As from the question,there are $n$ pairs of siblings.

The no. of way $n$ 'sibling couple' can be seated in the circle=$(n-1)$!

Each sibling can also sit together in $2$ different ways

The no. of ways we can make siblings sit together is=$2(n-1)$!

The no. of ways we cannot make siblings sit together is=$(2n-1)!-2(n-1)$!

This is the link of the graph

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    $\begingroup$ This is incorrect for several reasons. First, you’re trying to calculate the wrong thing: if this calculation were done correctly, it would give you the number of arrangements in which at least one sibling pair does not sit together. What is wanted is something quite different: the number of arrangements in which no sibling pair sits together. And it isn’t done quite correctly: the number of arrangements in which all $n$ sibling pairs are together is $2^n(n-1)!$, not $2(n-1)!$. $\endgroup$ – Brian M. Scott Oct 18 '16 at 13:36
  • $\begingroup$ @BrianM.Scott,Yes,you are right. $\endgroup$ – user237454 Oct 18 '16 at 13:39

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