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In a room with $k$ chairs in a circle, $m$ men and $w$ women take seats, where $m+w=k$. What is the probability that every man has a woman on his left and right?

kind of lost here.

José Carlos Santos
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Nikko
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1 Answers1

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To have a desired configuration, each man must have a woman on his right, so you can make $m$ man-woman pairs and $w-m$ single women. How many ways to distribute them out of how many ways without pairing?

Added:To pair them up line up the men and select women to pair with each. You can do that in $\frac {w!}{(w-m)!}$ ways. Now you have $w$ items (some pairs, some single women) to arrange around the circle, which you can do in $(w-1)!$ ways.f The overall probability is then $\frac {\frac {w!}{(w-m)!}(w-1)!}{(m+w-1)!}$

Ross Millikan
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  • not very clear for me, maybe if you write the combinatory would be more enlightening – Nikko Jun 19 '17 at 16:15
  • Can you do the number of ways to distribute the men and women without worrying about whether men are next to women? – Ross Millikan Jun 19 '17 at 17:10
  • absolutely. (m+w)!, no? – Nikko Jun 19 '17 at 19:01
  • Not quite. A circular table usually implies that there is nothing to designate any seat as special, so you need to divide by the number of seats. You can think of the first person to be seated as having only one choice of seat. – Ross Millikan Jun 19 '17 at 19:21
  • Yeah it should be like: (m+w-1)! – Nikko Jun 19 '17 at 19:24
  • To expand on my answer, clearly we must have every man having a woman on his right. If every man has a woman on his right, no man will have a man on his left, so we have an acceptable arrangement. Take each man and designate the woman who will be on his right. Then when you seat a men, immediately seat the woman you designated on his right. Finally you can seat the unattached women in the remaining seats. – Ross Millikan Jun 19 '17 at 19:25
  • If I take each man and assignate a woman, that means as if 4 persons are available to seat. How many ways can I distribute them. 3!. But also the pairs can be distributed 2!. and I'm stucked. can you please solve the thing, I am sure if I see it will be a lot easier to understand. By the way the answers must be one of these: 50%,60%,72%,75%. – Nikko Jun 19 '17 at 20:53
  • You can't have a numeric answer if we are not given $m$ and $w$. – Ross Millikan Jun 19 '17 at 21:09
  • true. m=2 w=4 total seats 6. – Nikko Jun 19 '17 at 21:25