discrete mathematics problem

Question

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.]

Answer 1

First we make all the 10 women stand in a line. Now there are 11 places in between the women and on the two sides where you can make this six men stand. We can choose $6$ places from these $11$ places in $\binom{11}{6}$ ways, and we can obviously rearrange among the men in $6!$ ways and among the women in $10!$ ways, without altering the order of men and women in the line. SO the number of such arrangements is $$10!6!\binom{11}{6}=\frac{10!6!11!}{6!5!}=\frac{10!11!}{5!}$$