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Combinations/Permutations of Seating at a Circular Table 
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#1
Mar2313, 04:39 PM

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1. The problem statement, all variables and given/known data
In how many ways can a group of 9 women and 6 men be seated at a circular table if no two men can be seated next to each other? 2. Relevant equations 3. The attempt at a solution I have come up with a solution, but I'm unsure if my reasoning is correct... First, I started with the 9 women. They can be arranged in 8! ways / 9 rotations = 4480 options. Then, I drew a picture of a 12person table, placing a man between each woman. This is where I think I may have done something wrong. The men can be placed in 5! ways, divided by 12 for rotations = 10 options. Finally, I drew a picture of the 15person table. The three remaining women can be placed in 36 different places, with three possibilities existing between each female/male pair. This can be calculated as 36 choose 3, divided by 15 rotations = 476 options. The final answer would be all three options multiplied together, so 4480(10)(476). As I said, I have a feeling that I did something wrong, so I would really appreciate some suggestions to get me back on track :) 


#2
Mar2313, 04:46 PM

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I would not consider rotations for men and women separately  the circular symmetry just applies to the whole table, not to both genders separately. You can divide the total number by 15 afterwards, this should be easier.
I would begin with the men/womencombinations first, placing 3 women somewhere between existing men/womenpairs, and care about the individual men/women and the circular table afterwards. 


#3
Mar2313, 04:51 PM

P: 15

Okay, so you're saying I should start with a 12person table, where the men and women can both be arranged in 5! ways, and then multiply this by 36 choose 3 for the three remaining women, and finally divide this by 15?
So, [(5!)(5!)(36 choose 3)] / 15? 


#4
Mar2313, 06:49 PM

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Combinations/Permutations of Seating at a Circular Table
Are the positions around the table distinct? I.e. if one arrangement is a rotation of another, are these counted separately?



#5
Mar2313, 07:00 PM

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@haruspex: I think "circular table" implies that the positions are not distinct.
@annpaulveal: Why 5!? You have 15 distinct chairs to fill (imagine a line of 15 chairs, for example), they will lose their identity afterwards. And I don't think your new idea works. 


#6
Mar2313, 07:52 PM

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There are two ways I can think of to do this:



#7
Mar2313, 09:39 PM

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#8
Mar2313, 09:44 PM

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I'm going to persist with the 'distinct positions' interpretation for now.
Try thinking of it in terms of which women get a man seated to their left. 


#9
Mar2413, 09:26 AM

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#10
Mar2413, 04:19 PM

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Fwiw, if seats were to be distinct, the answer is ##\left(^{w}_m\right)+\left(^{w1}_{m1}\right)##. 


#11
Mar2413, 06:25 PM

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It may help you to look at a case where it's easy to count by hand. For example, say you have two men and four women. You should be able to convince yourself the only ways you can arrange them are: mWmWWW mWWmWW mWWWmW Any other arrangement around a circular table would have the two men sitting next to each other. You don't need to include arrangements where a woman is very first because those arrangements are simply one of the above shifted cyclicly to the right. There are 2! ways to seat the men into the two m spots, and 4! ways to seat the women in the four W spots. In total, you have 2×24×3 ways to seat the group at the table without the men sitting next to each other. So how do you come up with 3, the number of arrangements? In terms of pairs, you can look at the arrangements as (MW)(MW)WW (MW)W(MW)W (MW)WW(MW) Can you come up with an expression that counts these? 


#12
Mar2513, 12:28 AM

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#13
Mar2513, 04:23 AM

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Oy. OK, I'll leave this problem to you guys.



#14
Mar2513, 12:50 PM

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Think that you place 9 women around the circular table. As the question requires the number of circular permutations, there are (91)! ways to place 9 women. Now there are 9 spaces to place 6 men such that none of these 6 men sit next to each other. Then, just find the number of ways to place 6 men into these 9 spaces. After that simply by using the product rule, you should be able to find the result



#15
Mar2613, 01:59 AM

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#16
Mar2613, 05:31 AM

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