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Five girls including Ramya and Five boys including Randhir are to be seated around a circular table such that no. two boys are adjacent and Ramya is not adjacent to Randhir. In how many ways this can be done?

My approach:-

All the girls can be seated on the chairs in $4!$ ways around the circular table.

Since no two boys are seated adjacent to each other, I took them to be seated on the chairs between two adjacent girls, since there are 5 seats available so the boys can occupy these 5 seats in $5!$ ways.

Now for the 2nd condition that Ramya and Randhir should not be seated together : Ramya can be seated on any of the 5 chairs which was initially selected for girls, correspondingly there will be 2 seats adjacent to her where Randhir would be seated.

so I am getting $4!*5!$ - $(5*2)$ ways , which is incorrect ! I am not able to understand the flaw in my approach, please help !

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  • $\begingroup$ Your exclusion is done in a very weird way. Are you saying that there are only 10 ways for R-R to seat together? That seems very few to me. So, can you elaborate further? $\endgroup$ Commented Oct 21, 2024 at 20:24
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    $\begingroup$ No need for exclusion here...arrange the $5$ boys however. That creates $5$ gaps each of which must contain exactly one girl. Declare that Randhir is at the head of the table, that establishes the rotation. Pick a gap for Ramya, there are only three that meet the condition. After that, you can assign freely. $\endgroup$ Commented Oct 21, 2024 at 20:30
  • $\begingroup$ @Calvin Lin Oh yes thats a very small number, I should take the arrangements of rest of the people as well... so maybe what I can do is first fix one R ,then rest of the girls can be arranged in 4! ways, also the second R has to be adjacent to this one so rest of the boys can be arranged in 4! ways, but the 2nd R can be on either side so 2*4!*4! should be deduced ? $\endgroup$ Commented Oct 21, 2024 at 20:31
  • $\begingroup$ thanks @lulu I got the answer by that approach :)) , but can you please help me with my method as well, does this exclusion which I have stated in above comment makes sense ? $\endgroup$ Commented Oct 21, 2024 at 20:35
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    $\begingroup$ If we assign Randhir seat $1$, and fix the positions of the boys ($4!$ ways to do that), then there are $4!$ arrangements in which Ramya is in gap $1$ and $4!$ arrangements in which she is in gap $5$, with what I hope is obvious notation. $\endgroup$ Commented Oct 21, 2024 at 20:38

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The simple way is to fix Randhir at "head" of table, fix the rest of the boys in $4!$ ways, fix Ramya in $3$ ways, and the rest of the girl in$4!$ ways, so $4!\times 3\times 4! = 1728$ ways

But if you want to apply inclusion-exclusion, (again with Ranadhir at head of table) fix the boys in $4!$ ways, fix the girls in $(5!- 2\times 4!) =1728$ [Unconstrained girls can sit in $5!$ ways, - $2$ wrong seats for Ramya $\times 4!$ for the rest ]

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  • $\begingroup$ This, of course takes seats as unnumbered as is the convention unless numbered seats are explicitly mentioned. $\endgroup$ Commented Oct 21, 2024 at 21:17

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