How many ways are there for 8 men and 5 women to stand in a line so that no two women stand next to each other? I have a homework problem in my textbook that has stumped me so far. There is a similar one to it that has not been assigned and has an answer in the back of the textbook. It reads: 

How many ways are there for $8$ men and $5$ women to stand in a line so that no two women stand next to each other?

The answer is $609638400$, but no matter what I try I cannot reach that number. I have tried doing $2(8!5!/3!)*(8!/5!)$ since each woman must be paired with a man in order to prevent two women getting near each other. But of course, it's the wrong answer. 

What am I doing wrong here?
For this type of problem first consider the position of the men, then the position of the women.In general first consider the the group with more people.

How many possible ways are there to arrange eight men in a row? It will be $ ^8P_8 = 8! = 40320$

Now as no two women stand next to each other,we can imagine the situation as 


                   * M * M * M * M * M * M * M * M *



Hence, we need to find how many ways we can arrange $5$ women in the $9$ possible (as shown above) places,this is actually $ ^9P_5 = 9*8*7*6*5 = 15120$

Now applying the fundamental law of counting (precisely product rule), total number of possible arrangements satisfying both constraints is: $15120 * 40320 = 609638400$ which is your required/desired answer.
M * M * M * M * M * M * M * M * .


there are 9 "*" .
8 "M"  .
now select 5 * from 9 * then == c(9,5)