How many ways are there to seat n married couples in a row of 2n chairs such that the couples never sit next to each other? Shouldnt it be as as simple as n.C.2n divided by 2!
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Using inclusion/exclusion $\displaystyle \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \binom{n}{k}\left( {2n - k} \right)!\left( {2^k } \right)} $.
Last edited by Plato; Oct 17th 2009 at 06:27 AM.
Originally Posted by Plato Using inclusion/exclusion $\displaystyle \sum\limits_{k = 0}^n {\left( { - 1} \right)^k \binom{n}{k}\left( {2n - k} \right)!\left( {2^k } \right)} $. @ Plato - Hi Did you actually workout all the cases - "1,2,3,...,n couple together" and took a compliment or there was an easy way? Thanks
Originally Posted by aman_cc Did you actually workout all the cases - "1,2,3,...,n couple together" and took a compliment Exactly. I do not know another way.
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