If 2n people are sitting at equally spaced intervals around a circle. How many ways can they form n pairs if no two persons seated directly opposite each other can form a pair?
What I did was draw some examples for low values of 'n' and also tried to see it as a graph on 2n vertices where there is no edge between opposite vertices. However, I have not come to a revelation. Any help would be appreciated.
Hint
$2n$ persons can form $\binom{2n}{2}$ pairs without restrictions.
How many pairs must we subtract from that due to the restriction ?
I am assuming that the $2n$ persons have already been seated in some order.
Then each of $n$ persons will have a specific person opposite with whom they can't form a pair,
thus $\binom{2n}{2} - n$ allowable pairs.
As shown at the beginning of this answer, without restriction, there are $(2n-1)!!$ ways to form pairs with $2n$ people. Let us count how many ways there are to pair at least one opposite using Inclusion-Exclusion.
There are $\binom{n}{k}$ ways to choose $k$ opposites. For each of those choices, there are $(2n-2k-1)!!$ ways to form pairs with the $2n-2k$ left. Inclusion-Exclusion says that there are $$ \sum_{k=1}^n(-1)^{k-1}\binom{n}{k}(2n-2k-1)!! $$ pairings with at least one pair of opposites. Subtracting this from $(2n-1)!!$ gives $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^n(-1)^k\binom{n}{k}(2n-2k-1)!!} $$ ways to pair without matching opposites. $$ \begin{array}{c|c} n&1&2&3&4&5&6&7&8&9\\\hline \text{pairings}&0&2&8&60&544&6040&79008&1190672&20314880 \end{array} $$ Here are the $8$ pairings for $n=3$:
