10 people are seated at a circular table for dinner. They get up and later return for a second course. How many ways can they be re-seated so that everybody is beside two new people?
My attempt at an answer was the following. WLOG assume that person $i$ has person $i+1$ to their left so that the first seating arrangement is like a clock with ten hours. We count instead the number of invalid combinations. Let $A_i$ be the event that person $i$ is "unhappy". Then $|\cup_i A_i|=\sum_i|A_i|-\sum_{i<j}|A_i\cap A_j|+\sum_{i<j<k}|A_i\cap A_j\cap A_k|-\ldots-|A_1\cap A_2\cap\ldots\cap A_{10}|$. Out of intersections/unions of happiness/unhappiness intersections of unhappiness are easiest to calculate, and I have calculated all $|A_i|$, $|A_i\cap A_j|$ and some $|A_i\cap A_j\cap A_k|$ but there are so many cases. See that for the triples there are six cases, $|A_i\cap A_{i+1}\cap A_{i+2}|$, $|A_i\cap A_{i+1}\cap A_{i+3}|$, $|A_i\cap A_{i+1}\cap A_{i+4}|$, $|A_i\cap A_{i+2}\cap A_{i+4}|$, $|A_i\cap A_{i+2}\cap A_{i+5}|$ and $|A_i\cap A_{i+3}\cap A_{i+6}|$, all others are equal to one of theses. For quadruples I can barely count how many cases there are.