text: ## how many ways-married couples
  
  For the discussion of math. Duh.
  
  Moderators: gmalivuk, Moderators General, Prelates
  
  evinda
  Posts: 32
  Joined: Fri May 03, 2013 10:16 pm UTC
  
  ### how many ways-married couples
  
  Hello!!!
  Could you help me at the following exercise?
  With how many ways can we choose a man and a woman that are not married to each other from n married couples?
  
  I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?
  
  Flumble
  Yes Man
  Posts: 2262
  Joined: Sun Aug 05, 2012 9:35 pm UTC
  
  ### Re: how many ways-married couples
  
  Can you explain how you got to (n-1)n? Or was it simply a hunch?
  
  Since #(choose unmarried man and woman) = #(choose man and woman) - #(choose married man and woman), I'd approach it from the right-hand side:
  Spoiler:
  there are n! ways to match the men to the women (a.k.a. there are n! permutations to sort the women keeping the men in place; you don't get different couples by shuffling the men) and there are n married couples, so there ought to be n!-n ways to pick a man and a woman who aren't married.
  
  On a sidenote: must all couples consist of a man and a woman?
  
   As gmalivuk pointed out I assumed all ways to pair up everyone instead of the ways to choose one couple.
  Spoiler:
  Which is quite easy, as you just have to choose one man and a woman he didn't marry (or vice-versa, doesn't matter), which makes for n*(n-1) possibilities.
   Or put differently: there are n*n ways to pick a male and a female, but n of those are the original couples.
  Last edited by Flumble on Fri Apr 25, 2014 7:14 pm UTC, edited 3 times in total.
  
  gmalivuk
  GNU Terry Pratchett
  Posts: 26820
  Joined: Wed Feb 28, 2007 6:02 pm UTC
  Location: Here and There
  Contact:
  
  ### Re: how many ways-married couples
  
  You seem to be counting the total possible ways to pair up all the men with all the women (excluding cases where a man is paired with his wife). The OP seems to be asking for a count of individual man+woman pairs.
  
  Though you're right that it is complicated by the fact that they haven't told us how many of the couples are gay.
  Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
  ---
  If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome
  
  (he/him/his)
  
  DavCrav
  Posts: 251
  Joined: Tue Aug 12, 2008 3:04 pm UTC
  Location: Oxford, UK
  
  ### Re: how many ways-married couples
  
  evinda wrote:Hello!!!
  Could you help me at the following exercise?
  With how many ways can we choose a man and a woman that are not married to each other from n married couples?
  
  I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?
  
  So firstly, how many ways are there of choosing a man and a woman? How many of those choices lead to a married couple? Subtract one from the other.
  
  Thesh
  Posts: 6598
  Joined: Tue Jan 12, 2010 1:55 am UTC
  
  ### Re: how many ways-married couples
  
  So is the question intended to be basically equivalent to the following?
  
  You have n pegs and n holes, each peg fits into only one of the holes. How many ways can you place the pegs next to the holes so that no peg is next to a hole it fits into?
  Summum ius, summa iniuria.
  
  somehow
  Posts: 99
  Joined: Wed Aug 14, 2013 8:31 pm UTC
  
  ### Re: how many ways-married couples
  
  No, I don't think that's equivalent. I think the OP is asking how many ways we can pick one peg and one hole such that the peg does not fit into the hole.
  
  Spoiler:
  Each man is a member of n-1 possible pairs consisting of a man and a woman not married to each other (one for each of the other n-1 couples besides his own). There are n men. This gives n*(n-1) = n2-n possible choices of one man and one woman such that they are not married. Each such pair contains exactly one of the n men and so will have been counted exactly once.
  Xenomortis wrote:O(n2) takes on new meaning when trying to find pairs of socks in the morning.
  
  patzer
  Posts: 432
  Joined: Fri Mar 30, 2012 5:48 pm UTC
  Contact:
  
  ### Re: how many ways-married couples
  
  Not enough information. How many gay married couples exist out of the total? Are all of the people either men or women, or are there some who are neither or both? How many of the people are polygamous?
  
  Assuming entirely heterosexual monogamous couples:
  
  Spoiler:
  There are n men and n women. For each of the n men there are n-1 women who are not his wife. So the total number is n(n-1), or n2-n.
  
  Assuming heterosexual and homosexual monogamous couples:
  
  Spoiler:
  Assume there are a male-male couples, b female-female couples, and c male-female couples. (a+b+c=n). There are 2a+c men and 2b+c women.
  
  There are (2a+c)(2b+c) ways to pair a male with a female; with c of these pairings being subtracted for being a heterosexual couple. For a total of 4ab2+2ac+2bc+c2-c.
  
  I'm going to stop here before creating a tangled labyrinth of variables.
  If it looks like a duck, and quacks like a duck, we have at least to consider the possibility that we have a small aquatic bird of the family Anatidae on our hands. –Douglas Adams
  
  ConMan
  Shepherd's Pie?
  Posts: 1690
  Joined: Tue Jan 01, 2008 11:56 am UTC
  Location: Beacon Alpha
  
  ### Re: how many ways-married couples
  
  If we're looking for ways to pair up men and women from a set of heterosexual monogamous couples such that no-one ends up with their spouse, then we're asking for the number of derangements of the set of women (or equivalently men). As for the number of ways to choose a man and then choose a woman who is not his wife, then it is n*(n-1).
  pollywog wrote:
  Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
  I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.
explanation: The document is a discussion thread from a math forum where a user named evinda asks a combinatorial question about choosing a man and a woman who are not married to each other from a set of n married couples. The user provides an answer and asks for confirmation. Other users, including Flumble, gmalivuk, and DavCrav, provide feedback and explanations.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## how many ways-married couples
  
  For the discussion of math. Duh.
  
  Moderators: gmalivuk, Moderators General, Prelates
  
  evinda
  Posts: 32
  Joined: Fri May 03, 2013 10:16 pm UTC
  
  ### how many ways-married couples
  
  Hello!!!
  Could you help me at the following exercise?
  With how many ways can we choose a man and a woman that are not married to each other from n married couples?
  
  I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?
  
  Flumble
  Yes Man
  Posts: 2262
  Joined: Sun Aug 05, 2012 9:35 pm UTC
  
  ### Re: how many ways-married couples
  
  Can you explain how you got to (n-1)n? Or was it simply a hunch?
  
  Since #(choose unmarried man and woman) = #(choose man and woman) - #(choose married man and woman), I'd approach it from the right-hand side:
  Spoiler:
  there are n! ways to match the men to the women (a.k.a. there are n! permutations to sort the women keeping the men in place; you don't get different couples by shuffling the men) and there are n married couples, so there ought to be n!-n ways to pick a man and a woman who aren't married.
  
  On a sidenote: must all couples consist of a man and a woman?
  
   As gmalivuk pointed out I assumed all ways to pair up everyone instead of the ways to choose one couple.
  Spoiler:
  Which is quite easy, as you just have to choose one man and a woman he didn't marry (or vice-versa, doesn't matter), which makes for n*(n-1) possibilities.
   Or put differently: there are n*n ways to pick a male and a female, but n of those are the original couples.
  Last edited by Flumble on Fri Apr 25, 2014 7:14 pm UTC, edited 3 times in total.
  
  gmalivuk
  GNU Terry Pratchett
  Posts: 26820
  Joined: Wed Feb 28, 2007 6:02 pm UTC
  Location: Here and There
  Contact:
  
  ### Re: how many ways-married couples
  
  You seem to be counting the total possible ways to pair up all the men with all the women (excluding cases where a man is paired with his wife). The OP seems to be asking for a count of individual man+woman pairs.
  
  Though you're right that it is complicated by the fact that they haven't told us how many of the couples are gay.
  Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
  ---
  If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome
  
  (he/him/his)
  
  DavCrav
  Posts: 251
  Joined: Tue Aug 12, 2008 3:04 pm UTC
  Location: Oxford, UK
  
  ### Re: how many ways-married couples
  
  evinda wrote:Hello!!!
  Could you help me at the following exercise?
  With how many ways can we choose a man and a woman that are not married to each other from n married couples?
  
  I thought that it is (n-1)n ,but I am not sure.Can you tell me if it is right?
  
  So firstly, how many ways are there of choosing a man and a woman? How many of those choices lead to a married couple? Subtract one from the other.
  
  Thesh
  Posts: 6598
  Joined: Tue Jan 12, 2010 1:55 am UTC
  
  ### Re: how many ways-married couples
  
  So is the question intended to be basically equivalent to the following?
  
  You have n pegs and n holes, each peg fits into only one of the holes. How many ways can you place the pegs next to the holes so that no peg is next to a hole it fits into?
  Summum ius, summa iniuria.
  
  somehow
  Posts: 99
  Joined: Wed Aug 14, 2013 8:31 pm UTC
  
  ### Re: how many ways-married couples
  
  No, I don't think that's equivalent. I think the OP is asking how many ways we can pick one peg and one hole such that the peg does not fit into the hole.
  
  Spoiler:
  Each man is a member of n-1 possible pairs consisting of a man and a woman not married to each other (one for each of the other n-1 couples besides his own). There are n men. This gives n*(n-1) = n2-n possible choices of one man and one woman such that they are not married. Each such pair contains exactly one of the n men and so will have been counted exactly once.
  Xenomortis wrote:O(n2) takes on new meaning when trying to find pairs of socks in the morning.
  
  patzer
  Posts: 432
  Joined: Fri Mar 30, 2012 5:48 pm UTC
  Contact:
  
  ### Re: how many ways-married couples
  
  Not enough information. How many gay married couples exist out of the total? Are all of the people either men or women, or are there some who are neither or both? How many of the people are polygamous?
  
  Assuming entirely heterosexual monogamous couples:
  
  Spoiler:
  There are n men and n women. For each of the n men there are n-1 women who are not his wife. So the total number is n(n-1), or n2-n.
  
  Assuming heterosexual and homosexual monogamous couples:
  
  Spoiler:
  Assume there are a male-male couples, b female-female couples, and c male-female couples. (a+b+c=n). There are 2a+c men and 2b+c women.
  
  There are (2a+c)(2b+c) ways to pair a male with a female; with c of these pairings being subtracted for being a heterosexual couple. For a total of 4ab2+2ac+2bc+c2-c.
  
  I'm going to stop here before creating a tangled labyrinth of variables.
  If it looks like a duck, and quacks like a duck, we have at least to consider the possibility that we have a small aquatic bird of the family Anatidae on our hands. –Douglas Adams
  
  ConMan
  Shepherd's Pie?
  Posts: 1690
  Joined: Tue Jan 01, 2008 11:56 am UTC
  Location: Beacon Alpha
  
  ### Re: how many ways-married couples
  
  If we're looking for ways to pair up men and women from a set of heterosexual monogamous couples such that no-one ends up with their spouse, then we're asking for the number of derangements of the set of women (or equivalently men). As for the number of ways to choose a man and then choose a woman who is not his wife, then it is n*(n-1).
  pollywog wrote:
  Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
  I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a math forum where a user named evinda asks a combinatorial question about choosing a man and a woman who are not married to each other from a set of n married couples. The user provides an answer and asks for confirmation. Other users, including Flumble, gmalivuk, and DavCrav, provide feedback and explanations.
  
  Label: question-and-answer