In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats? In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats?
Please use principle of inclusion exclusion to solve.
My approach:
Let $S_i$ represent the number of arrangements where $i$ men are adjacent. Thus, $S_2$, for instance, would be ${4 \choose 2}(2!)(6!)$, treating the two adjacent men as one person in the circular arrangement (leaving seven people so $6!$ ways), and there are $2!$ ways to arrange these two adjacent men.
The whole equation would look like:
$$7! -  S_2 + S_3 - S_4 = 7! - {4 \choose 2}(2!)(6!) + {4 \choose 3}(3!)(5!) - {4 \choose 4}(4!)(4!) = -1296$$
I am not sure what I did wrong as the result should be positive.

You're overcounting. Take $S_2$: after picking the 2 men AB that are adjacent, you allow everyone else to be placed in $6!$ ways. However, that will include ways in which the other 2 men CD are adjacent as well. So, you are double counting ways in which AB are adjacent and CD are adjacent. 
I'll leave it to you to correct your formulas to account for those over-counts.  However, please that with the same number of men and women, they all need to alternate if you want to avoid placing two men adjacent to each other, meaning that starting with 1 woman, you have $3!$ ways to place the other $3$ women, and $4!$ ways to place the $4$ men. So, the total should come out to $3!4!$.  And for $n$, it should be $(n-1)!n!$
