6 men and 3 women stand in a Straight line
How many diff ways can they stand such that no two women stand next to each other?
The men can arrange themselves in $\displaystyle 6!$ ways in a line. Then women can stand in the $\displaystyle 5$ gaps between the men as well as at the $\displaystyle 2$ extreme ends of the line, so they can arrange themselves in $\displaystyle {\rm^7P_3}=7\times6\times5$ ways. Hence the total number of ways of arranging the $\displaystyle 9$ people is $\displaystyle 7\times6\times5\times6!=151\,200.$