If 10 married couples are seated at random on a round table , then what could be the probability of no wife get seated next to her husband .
I thought of making combination . But not able to do that or anything else .
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My comment on an answer to another question might suggest something like $\dfrac{ 2^0{5 \choose 0} 9! - 2^1{5 \choose 1} 8! + 2^2{5 \choose 2} 7! - 2^3{5 \choose 3} 6! + 2^4{5 \choose 4} 5!- 2^3{5 \choose 5} 4! }{9!}$ for a round table – Henry Oct 17 '16 at 17:22
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@Henry could you please explain me each term – ankit gupta Oct 17 '16 at 18:11
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@Henry there are 10 married couples not 5 married couples – ankit gupta Oct 17 '16 at 18:17
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Fair enough. For $n=10$ use inclusion-exclusion to calculate $\displaystyle \dfrac{1}{(2n-1)!}{\sum_{m=0}^n (-2)^m {n \choose m}(2n-1-m)!}$ which is about $0.3395$. The limit as $n$ increases is likely to be close to $e^{-1}\approx 0.367879$ – Henry Oct 17 '16 at 19:27