Combinatorics: $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together.











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$N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?




This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!



It is not necessary that women and men alternate, ACaBcb










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    up vote
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    down vote

    favorite













    $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?




    This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!



    It is not necessary that women and men alternate, ACaBcb










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?




      This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!



      It is not necessary that women and men alternate, ACaBcb










      share|cite|improve this question
















      $N$ couples sitting in a row, $M$ couples in the $N$ couples can't sit together. How many ways are there to arrange the seat?




      This is a variation of ménage problem, I thought I could arrange the M couples first using ménage problem's solution(https://www.math.dartmouth.edu//~doyle/docs/menage/menage/menage.html), but then it does not count ACaBcb, where $M=2$ (that is Aa, Bb) and $N=3$. Thanks!



      It is not necessary that women and men alternate, ACaBcb







      combinatorics






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      edited Nov 18 at 7:56









      Robert Z

      91.6k1058129




      91.6k1058129










      asked Nov 18 at 7:11









      kevin

      1056




      1056






















          1 Answer
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          Hint. By the inclusion-exclusion principle we have
          $$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
          Can you take it from here?






          share|cite|improve this answer























          • Thanks! Alternate till M couples right?
            – kevin
            Nov 18 at 7:38






          • 1




            @kevin Yes, that's correct.
            – Robert Z
            Nov 18 at 7:39










          • Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
            – kevin
            Nov 18 at 8:29








          • 1




            No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
            – Robert Z
            Nov 18 at 8:44












          • Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
            – kevin
            Nov 18 at 9:22











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          up vote
          1
          down vote



          accepted










          Hint. By the inclusion-exclusion principle we have
          $$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
          Can you take it from here?






          share|cite|improve this answer























          • Thanks! Alternate till M couples right?
            – kevin
            Nov 18 at 7:38






          • 1




            @kevin Yes, that's correct.
            – Robert Z
            Nov 18 at 7:39










          • Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
            – kevin
            Nov 18 at 8:29








          • 1




            No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
            – Robert Z
            Nov 18 at 8:44












          • Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
            – kevin
            Nov 18 at 9:22















          up vote
          1
          down vote



          accepted










          Hint. By the inclusion-exclusion principle we have
          $$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
          Can you take it from here?






          share|cite|improve this answer























          • Thanks! Alternate till M couples right?
            – kevin
            Nov 18 at 7:38






          • 1




            @kevin Yes, that's correct.
            – Robert Z
            Nov 18 at 7:39










          • Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
            – kevin
            Nov 18 at 8:29








          • 1




            No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
            – Robert Z
            Nov 18 at 8:44












          • Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
            – kevin
            Nov 18 at 9:22













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint. By the inclusion-exclusion principle we have
          $$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
          Can you take it from here?






          share|cite|improve this answer














          Hint. By the inclusion-exclusion principle we have
          $$(2N)!-binom{M}{1}cdotunderbrace{2binom{2N-1}{1}1!(2N-2)!}_{text{at least one of the $M$ couples together}}+binom{M}{2}cdotunderbrace{2^2binom{2N-2}{2}2!(2N-4)!}_{text{at least two of the $M$ couples together}}-dots$$
          Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 18 at 8:47

























          answered Nov 18 at 7:31









          Robert Z

          91.6k1058129




          91.6k1058129












          • Thanks! Alternate till M couples right?
            – kevin
            Nov 18 at 7:38






          • 1




            @kevin Yes, that's correct.
            – Robert Z
            Nov 18 at 7:39










          • Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
            – kevin
            Nov 18 at 8:29








          • 1




            No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
            – Robert Z
            Nov 18 at 8:44












          • Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
            – kevin
            Nov 18 at 9:22


















          • Thanks! Alternate till M couples right?
            – kevin
            Nov 18 at 7:38






          • 1




            @kevin Yes, that's correct.
            – Robert Z
            Nov 18 at 7:39










          • Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
            – kevin
            Nov 18 at 8:29








          • 1




            No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
            – Robert Z
            Nov 18 at 8:44












          • Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
            – kevin
            Nov 18 at 9:22
















          Thanks! Alternate till M couples right?
          – kevin
          Nov 18 at 7:38




          Thanks! Alternate till M couples right?
          – kevin
          Nov 18 at 7:38




          1




          1




          @kevin Yes, that's correct.
          – Robert Z
          Nov 18 at 7:39




          @kevin Yes, that's correct.
          – Robert Z
          Nov 18 at 7:39












          Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
          – kevin
          Nov 18 at 8:29






          Hi Prof, from a logical point of view, isn't the complement of "At least $1$ of the $M$ couples together" equal to "None of the $M$ couples together"? Suppose that's true, why alternate with additional terms? Thanks!
          – kevin
          Nov 18 at 8:29






          1




          1




          No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
          – Robert Z
          Nov 18 at 8:44






          No, "at least 1 of the M couples together" means a particular couple of the M couples together multiplied by $M$. Revise the inclusion-exclusion principle: en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
          – Robert Z
          Nov 18 at 8:44














          Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
          – kevin
          Nov 18 at 9:22




          Hi Prof, thanks! Now I see that $A_i$ stands for the event that the $i^{th}$ couple sits together.
          – kevin
          Nov 18 at 9:22


















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