There are a lot of similar questions to this but none that is quite the same so I figured I would ask a new question. The problem is you have a group of people that came in pairs, in how many ways can the $N$ people be arranged in a circle such that no person in the circle is holding hands with the partner they came with.
I have worked out by hand that for two couples(four people) the answer is $2$ ways
For three couples ($6$ people) the answer is $32$ ways
Also it should be noted that a rotated circle is not counted as a different arrangement.
If the answer posted below is correct the OEIS sequence is http://oeis.org/A129348 Can any one else confirm that this is correct?
Best Answer
Let m be the number of pairs, so that $n=2m$.
Let S be the set of all seatings of the pairs in a circle, and
let $E_i$ be the set of seatings in which pair i is together, for $1\le i\le m$.
Using Inclusion-Exclusion and the fact that n people can be seated in a circle in $(n-1)!$ ways,
$\displaystyle|\overline{E_1}\cap\cdots\cap\overline{E_m}|=|S|-\sum_{i}|E_i|+\sum_{i<j}|E_i\cap E_j|-\sum_{i<j<k}|E_i\cap E_j\cap E_k|+\cdots+(-1)^m|E_1\cap\cdots\cap E_m|$
$\;=(2m-1)!-\binom{m}{1}2^1(2m-2)!+\binom{m}{2}2^2(2m-3)!-\binom{m}{3}2^3(2m-4)!+\cdots+(-1)^m\binom{m}{m}2^m(m-1)!$
$\;\;=\displaystyle\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}2^{i}(2m-i-1)!$.