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**Plato** I will help you with circular arrangements.

Seat one of the non-married people at the table.

Now there are $\displaystyle 15!$ ways to seat the remaining people without any restrictions.

Now we need to find out how many ways to arrange these fifteen people so that at least one married couple are seated together, there are $\displaystyle \sum\limits_{k=1}^3 {\left( { - 1} \right)^{k+1}\binom{3}{k} \left( {2^k } \right)\left[ {\left( {15 - k} \right)!} \right]}$ ways to do that. . That is an inclusion/exclusion solution.

The idea is that we can use the one person already seated at the table as a reference point.