Firstly, we'll fix the postions where the [tex]n[/tex] women will be. We have [tex]n![/tex] forms to do that. So, we'll obtain a row like:
[tex]\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}[/tex]
The n+1 spaces represented by the underline positions will receive the men of the row. Then,
[tex]x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)[/tex]
Since there is no women sitting together, we must write that [tex]x_2,x_3,...,x_{n-1},x_n\ge1[/tex]. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:
[tex]\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}[/tex]
The equation (i) can be rewritten as:
[tex]x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\
x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\
x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)[/tex]
We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: [tex]\dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}[/tex]
[I can write the proof if you want]
Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: [tex]m![/tex]
Multiplying all results:
[tex]n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\
\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}[/tex]
