$m$ men and $w$ women are to be seated in a row so that no two women sit together. If $m > w,$ then the number of ways in which they can be seated is:

We first arrange the m men. This can be done in $m!$ ways. After m men have taken their seats, the women must choose w seats out of $(m + 1)$ seats marked with X  below.

$XMXMXMX\dots XMX$

$\text{ 1st 2nd 3rd }m\text{ th }(m+1)\text{ th }$

They can choose $w$ seats in ${ }^{m+1}{C}_w$ ways and arrange w women in $w!$ ways.

Thus, the required number of arrangements is

$m!\left({ }^{m+1}{C}_w\right)(w!)=\frac{m!(m+1)!w!}{w!(m+1-w)!}=\frac{m!(m+1)!}{(m+1-w)!}$