TITLE: How many ways are there for 10 women and six men to stand in a line
QUESTION [7 upvotes]: How many ways are there for 10 women and six men
to stand in a line so that no two men stand next to each
other? [Hint: First position the women and then consider
possible positions for the men.]

REPLY [11 votes]: As your hint suggests, for this sort of problem, first consider the positions of the women, then the positions of the men. 
How many possible ways are there to arrange ten women in a row? It will be $$^{10}P_{10}=10! = 3628800.$$
Now since we need for no two men to stand next to one another, we can visualize the situation as follows

                   * W * W * W * W * W * W * W * W * W * W *


A man can be inserted in any one of the eleven positions marked off with a '*', and this will ensure no two men stand next to each other. Hence, we need to find how many ways we can arrange $6$ men in the $11$ possible places: this is given by $$ ^{11}P_6 = 11\times 10\times 9\times 8 \times 7\times 6 = 332640.$$
Now applying a fundamental law of counting (specifically, the rule of the product), to calculate the total number of possible arrangements satisfying both constraints. This gives us : $$3628800\times 332640 = 1,207,084,032,000$$ which is your required/desired answer.