What is wrong with my solution for a combinatorics problem? The problem is:

We want to place m men and n women at a round table with $\left(m+n\right)$ chairs in such a way that no two men sit next to each other. In how many ways can this be done?

I approached the problem in the following way. Knowing that between each man there have to be at least one woman and they are sitting at the round table, then we are left with $\left(n-m\right)$ women to arrange. There are exactly $m$ places (gaps, so $m-1$ walls dividing these places) to put women, there are $P_0\left(n-m, m-1\right)=\frac{(n-m+m-1)!}{\left(n-m\right)!\left(m-1\right)!}=\frac{\left(n-1\right)!}{\left(n-m\right)!\left(m-1\right)!}$ ways to arrange women in these places (arrangement with repetitions of elements with 2 types). Then as on each chair sits only either man or woman, I multiplied the answer by $2$ (it is like shifting all combinations by 1 chair). So $2m!n!\frac{\left(n-1\right)!}{\left(n-m\right)!\left(m-1\right)!}$ 
But the real answer is $m!n!\left[\frac{\left(n-1\right)!}{\left(n-m\right)!\left(m-1\right)!} + \frac{n!}{\left(n-m\right)!m!}\right]$. Explaining that we fix one chair and consider 2 cases, when here sits man, and then woman. This solution is clear, and I got it. But I can't anyway understand why my solution fails. Rather, I can't understand how to "see" the second case from my approach. What I can see is that I need to shift, but I don't find fixing a chair necessary. I will really appreciate if you at least show which cases I don't consider or instead calculate twice, etc. Thanks in advance!
 A: Lets assume that the seats are numbered clockwise $1,2,…,n+m$ then the count you are trying to double 
$$\frac{(n-1)!}{(n-m)!(m-1)!}$$
is the number of configurations with a man in seat $1$ and women in both seats $2$ and $n+m$. 
By rotating each such configuration $1$ seat clockwise the man previously in seat $1$ is now in seat $2$ and the two women are now in seats $1$ and $3$. 
Your claim is that adding this case accounts for all cases where a woman occupies seat $1$, however it only accounts for the cases where there is a woman in seat $1$ and a man in seat $2$. It ignores cases where there is a woman in seat $1$ and a woman in seat $2$. i.e. It undercounts.
A: I have a different view:
Your solution does not answer the question.  Let me explain the solution with a different perspective.
Since it is a round configuration, the way you bring it to a line configuration is to fix one chair with either a man or woman.
Suppose you fix a chair seated by a woman, the configuration would stand as below
W _ _ _ _.............. _. The spaces can be filled with any order and now that we want to fill $m$ men around $n-1$ women, we have a total number of arrangements would be(with the $m$ men and $n$ women could be permuted by $m!$ and $n!$)
${(n-1+1)\choose m}. m! n!$
Suppose you fix a chair seated by a man, the configuration would stand as below
M W _ _ _ _ ........._ _W. The spaces can be filled with any order and now that we want to fill $m-1$ men around $n-2$ women, we have a total number of arrangements would be(with the $m$ men and $n$ women could be permuted by $m!$ and $n!$)
${(n-2+1)\choose (m-1)}. m! n! = {(n-1)\choose (m-1)}. m! n!$
Sum of these should get you the answer.
Good luck
