If it were the three women that had to sit together i could solve this as by putting them in a single group among the men. If i multiply the permutation of these groups by the permutation of women inside this group i would get a total of 5! * 3! = 720.
I thought of using this strategy for the problem of arranging the men and multiplying by C(4,3)=4 because the group is a combination of 3 out of 4 men, which would give 4 * 720 = 2880.
From creating a list of all permutations, labelling them man/women and searching for three adjacent men i know however that the answer must be 2304. That means my approach above is double counting some of them.
For example when i would number the men and women these two permutations with the constructed group between brackets, would be equal:
(M1, M2, M3), M4, W1,W2, W3
M1, (M2, M3, M4), W1,W2, W3
Also from the result of 2304 i would suspect it to contain factors like 3! * 4! * 4 * 4 = 2304, but can't put it together.
Any help would be greatly appreciated.
Best Answer
I assume that you have four men, three women, each person is distinct, and we are wanting to count the number of ways that they can line up such that there are at least three men in a row (i.e. at least one man who has another man to both his left and his right).
I'm sure there are other ways to see the solution, but here I'll showcase one of them.
One of two things will happen, either all four men will be in a group together, or there will be at least one woman separating the men.
Arrange the women in a row ignoring the men for now ($3!$ options). Then, arrange the men in a row, ignoring the women ($4!$ options). Now, break into cases based on if there is going to be a group of all four men together or not and then interleave the men into slots between the women maintaining their order. If the men were all together, this could be done in $4$ ways, picking which slot between women they stand. If the men were split into a group of three and a group of one, then pick which slot was used for the group of three and then which is used for the group of one ($4$ options in the first case, $4\times 3$ options in the second case).
This gives us a total number of arrangements of $3!\times 4!\times (4 + 4\times 3)$
We could similarly have removed the "bad" arrangements where the men are all apart by splitting into cases similarly, but in this case it seems like more work rather than less. It would give $7! - 4!3!(\binom{4}{4} + 4\times \binom{3}{2} + \binom{4}{2})$ which should come to the same value as before.