In how many ways 5 couples can be seated around a circular table with some conditions attached? In how many ways 5 couples can be seated around a circular table such that men and women sit alternatively and no person sits adjacent to his or her spouse ?
Edit: The chairs are alike! 
I have gone through few answers to the question where 5 couples are arranged in such a way that they don't sit together using inclusion and exclusion principle but I am not getting my way around when men and women also have to sit alternatively!
 A: There are $4!$ ways to place the men in such a way that men and women will sit alternatively. 
Place them and label the spots where the men have taken their seats clockwise with $a,b,c,d,e$. 
Now we will take a look at the possible configurations for women.
Without further conditions there are $5!$ configurations for women.
Let $A$ denote the set of these configurations where the man that sits
at $a$ has his wife next to him.
This similar for $B,C,D,E$ where the capitals correspond with labels
$b,c,d,e$ respectively.
The answer to the question is then $4!\left|A^{\complement}\cap B^{\complement}\cap C^{\complement}\cap D^{\complement}\cap E^{\complement}\right|=4!\left(5!-\left|A\cup B\cup C\cup D\cup E\right|\right)$
so it remains to find $\left|A\cup B\cup C\cup D\cup E\right|$.
This can be done by means of inclusion/exclusion. Up to a certain
level we can also make use of symmetry (e.g. notice that of course
$\left|A\cap B\right|=\left|B\cap C\right|$) but here we must be
careful.
At first hand we find that: $$\left|A\cup B\cup C\cup D\cup E\right|=5\left|A\right|-5\left|A\cap B\right|-5\left|A\cap C\right|+5\left|A\cap B\cap C\right|+5\left|A\cap B\cap D\right|-5\left|A\cap B\cap C\cap D\right|+\left|A\cap B\cap C\cap D\cap E\right|$$
Then checking the cases one by one we find:


*

*$\left|A\right|=2\times4!=48$

*$\left|A\cap B\right|=3\times3!=18$

*$\left|A\cap C\right|=4\times3!=24$

*$\left|A\cap B\cap C\right|=4\times2!=8$

*$\left|A\cap B\cap D\right|=6\times2!=12$

*$\left|A\cap B\cap C\cap D\right|=5\times1!=5$

*$\left|A\cap B\cap C\cap D\cap E\right|=2\times0!=2$
So our final answer is: $$4!\left(5!-5\times48+5\times18+5\times24-5\times8-5\times12+5\times5-26\right)=24\times13=312$$

I hope that I did not make any mistakes. Check me on it.
A: There are $P(5)=(5-1)!=4!$ circular permuations of the $5$ men.
Now consider the arrangement: $*M_1*M_2*M_3*M_4*M_5$.
There are $3$ arrangements of $W_1$:
$$1) *M_1*M_2\color{red}{W_1}M_3*M_4*M_5\\
2) *M_1*M_2*M_3\color{red}{W_1}M_4*M_5\\
3) *M_1*M_2*M_3*M_4\color{red}{W_1}M_5\\$$
There are $3,2$ and $2$ arrangements of $W_2$, respectively:
$$\begin{align} 1) \ \ \ \ \ &I) \color{blue}{W_2}M_1*M_2\color{red}{W_1}M_3*M_4*M_5\\
\ \ \ \ \ \ \ \ &II) *M_1*M_2\color{red}{W_1}M_3\color{blue}{W_2}M_4*M_5\\
\ \ \ \ \ \ &III)*M_1*M_2\color{red}{W_1}M_3*M_4\color{blue}{W_2}M_5\\
2) \ \ \ \ \ &I)\color{blue}{W_2}M_1*M_2*M_3\color{red}{W_1}M_4*M_5\\
&II)*M_1*M_2*M_3\color{red}{W_1}M_4\color{blue}{W_2}M_5\\
3) \ \ \ \ \ &I)\color{blue}{W_2}M_1*M_2*M_3*M_4\color{red}{W_1}M_5\\
&II)*M_1*M_2*M_3\color{blue}{W_2}M_4\color{red}{W_1}M_5\\
\end{align}$$
There are $1,3,2,1,2,1,3$ arrangements of the rest women, respectively (verification is left as an exercise).
Due to symmetry, the final number of arrangements of the $5$ pairs is $4!\cdot 13=312$.
