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In how many ways can we seat $5$ pairs of twins in a row of $10$ chairs such that nobody sits next to his or her twin ?
To count the number of possible arrangements with PIE it's simple to calculate the no. of ways in which we can seat $1$ pair of twins to be $9!.2.5$ but how to calculate the the ways in which we can seat more than $1$ pair of twins? For $2$ pair of twins it is ${5\choose{2}}.2^2.8!$ but it's getting messy for more pairs so is there any general way.

Martin Sleziak
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sanemi678
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  • What is PIE ?... – Jean Marie Mar 29 '23 at 08:16
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    https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – Anne Bauval Mar 29 '23 at 08:17
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    @JeanMarie PIE=principle of inclusion and exclusion – callculus42 Mar 29 '23 at 08:30
  • $$10! + \sum_{i=1}^5 \left[ ~(-1)^i ~\binom{5}{i} 2^i ~(10-i)! ~\right].$$ $~\displaystyle \binom{5}{i} ~$ reflects picking which $~i~$ pairs will violate the constraint. $~2^i~$ reflects internally permuting the $~i~$ pairs. $~(10-i)! ~$ reflects permuting the $~(10-i)~$ external units. See this answer for an explanation of and justification for the Inclusion-Exclusion formula. – user2661923 Mar 29 '23 at 08:44

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