If $m$ and $n$ women are standing toghter ]such that no men are woman are adjacent together what are the number of Permutations

Question

suppose $m$ men and $n$ women from a single line in such a way that no two men are next to each other and no women are next to each other how many lineup are possible ?

Never solved these problems before with $m$ and $n$

Archaick · Answer 1 · 2015-04-18T07:29:00.627

1

Hint: this is only possible if $m=n$ or $n \pm 1$

edited Apr 18 '15 at 07:29

answered Apr 18 '15 at 07:14

Archaick

2,273

drhab · Answer 2 · 2015-04-18T07:24:56.533

1

If my understanding of your question is okay then no lineups are possible if $|m-n|>1$.

If $|m-n|=1$ then there are $m!n!$ possibilities.

If $m=n$ then there are $2m!n!=2n!n!=2m!m!$ possibilities. Factor $2$ because the person most left can be a man or a woman.

edited Apr 18 '15 at 07:24

answered Apr 18 '15 at 07:18

drhab

151,093

I agree this answer but I wonder why diffrent people are giving different answers – Connor Verlekar Apr 18 '15 at 07:26
No, this is correct. – Archaick Apr 18 '15 at 07:29
I think the answers are the same now. – drhab Apr 18 '15 at 07:29

If $m$ and $n$ women are standing toghter ]such that no men are woman are adjacent together what are the number of Permutations

2 Answers2