SOLUTION: Suppose you have a group of six men and three women; you want to line them up so that no two women are standing next to each other. In how many ways can this be done? I know the

Algebra ->  Permutations -> SOLUTION: Suppose you have a group of six men and three women; you want to line them up so that no two women are standing next to each other. In how many ways can this be done? I know the      Log On


   



Question 404758: Suppose you have a group of six men and three women; you want to line them up so that no two women are standing next to each other. In how many ways can this be done?
I know there has to be an easier way to figure this out than to just write M/F in combination until all of the combination's are exhausted (which is the method my classmates have done, and they have answers ranging from 12 to 30 ways) - we just have not gone over the actual formulas for this, and I can't figure it out from the textbook.

Answer by sudhanshu_kmr(1152) About Me  (Show Source):
You can put this solution on YOUR website!

no. of ways to arrange 6 men = 6P6 = 6!
# M # M # M # M # M # M #
any women can be put at place # .
total no. of place= 7
no. of ways to put 3 women at 7 places = 7P3 = 210

total no. of ways to arrange 6 men and 3 women = 151200

here the concept is :
there is no restriction with men so firstly arrange men and after arranging them
at N places, arrange the women at (N+1)places....