4. In how many ways can 6 men and 8 women be lined up such that men are not adjacent?
Since no men can be adjacent we know that each pair of men is separated by at least one woman as shown:
M W M W M W M W M W M (W W W remaining to be placed)
And, since the remaining three women can be placed anywhere into our existing arrangement, we have 36 possible spaces for them to be placed as shown:
_ _ _ M _ _ _ W _ _ _ M _ _ _ W _ _ _ M _ _ _ W _ _ _ M _ _ _ W _ _ _ M _ _ _ W _ _ _ M _ _ _
So we need to place 3 indistinguishable women (the women are identical, they have no labels, W1 W2 W3 = W3 W2 W1, order doesn't matter) into 36 spaces.
To do this we can use a strategy found in one of the previous problems, find the number of 3-permutation in a set of 36 elements:
P(36,3)= 36*35*34=42840
BUT, we have to be careful because the women are indistinguishable, and therefore we have over-count by a factor of the number of ways to label 3 objects (6).
So, we take P(36,3)/6 = 42840/6 = 7140 ways to arrange 6 men and 8 women such that no men are adjacent.
Note that finding the 3-permutations of a set with 36 elements and "removing the labels" (dividing by 6 in this case) is the same as using the "choose" system as shown:
C(36,3)= 36!/3!*33! = (34*35*36)/(1*2*3) = (34*35*36)/6 = 7140
As always, please let me know if any mistake has been made. --Msstaffo 22:45, 9 March 2009 (UTC)
