Assume I want to place N women and M men around a table with N+M seats such that no person(man or woman) may be seated in-between two women. I want to show that this is impossible when N>M. I assume this can be proven by the pigeonhole Principle? But how do I do this?
Placing men and women around a round table
2
$\begingroup$
combinatorics
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0Not sure about pigeonhole, but it looks like induction (on the total number of people) might work. After all, in your configuration there either is or is not a group of $n$ men next to each other for $n≥3$. If there is, then you can remove one of them without violating the rule. If there isn't then it is pretty easy to see that your condition implies $N≤M$. – 2017-02-19
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0Actually, you don't even need the induction. Juust note that we must alternate groups like $W^a,M^b$ where $a\in \{1,2\}$ and $b≥2$. Thus, pairing each male group with the clockwise next female group we get $N≤M$ pairing by pairing. – 2017-02-19
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0I don't understand there can't be more then three women next to each other and after every woman or pair of women there must be at least two men, so you can't have all the men next to each other – 2017-02-19
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0Not sure if that was a response to me or not. In any case, you actually can't have more than $2$ women next to each other, and I don't see the problem with having all the men next to each other (assuming there are no more than $2$ women of course). – 2017-02-19