In how many ways can $7$men and $7$women can be seated around a round table such that no two women can sit together? 

Given : $7$men and $7$women seated around a round table

To find: no of ways in which $7$men and $7$women can be seated around a round table such that no two women can sit together. 

A permutation is a (possible) rearrangement of objects. 

By using formula $n! = 1. 2. 3. .......(n-1). n$

$7$men can seat themselves in a round table in $(7-1)! = 6!$ ways . 

Now, there are $7$places vacant between these $7$men. 

$\therefore 7$ women can seat themselves in these  $7$places in $7!$ ways. 

$\therefore$ Total number of required arrangement where no two women sit together $= 6! \times 7!$