Combinatorics- distribute n objects to k bin with constraints












0












$begingroup$


Suppose I have n objects distributed to k bins and I want to know how many multisets satisfy the following condition: exactly j bins have at least m objects, where j $leq$ k,m $leq$ n, and j*m $leq$ n. For example, if n = 5, k=4, j=2 and m = 2, there are 24 subsets:



[3, 2, 0, 0]
[3, 0, 2, 0]
[3, 0, 0, 2]
[2, 3, 0, 0]
[2, 2, 1, 0]
[2, 2, 0, 1]
[2, 1, 2, 0]
[2, 1, 0, 2]
[2, 0, 3, 0]
[2, 0, 2, 1]
[2, 0, 1, 2]
[2, 0, 0, 3]
[1, 2, 2, 0]
[1, 2, 0, 2]
[1, 0, 2, 2]
[0, 3, 2, 0]
[0, 3, 0, 2]
[0, 2, 3, 0]
[0, 2, 2, 1]
[0, 2, 1, 2]
[0, 2, 0, 3]
[0, 1, 2, 2]
[0, 0, 3, 2]
[0, 0, 2, 3]



I'm having difficulty finding a general formula for this problem. Any help would be appreciated.






combinatorics







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose I have n objects distributed to k bins and I want to know how many multisets satisfy the following condition: exactly j bins have at least m objects, where j $leq$ k,m $leq$ n, and j*m $leq$ n. For example, if n = 5, k=4, j=2 and m = 2, there are 24 subsets:



    [3, 2, 0, 0]
    [3, 0, 2, 0]
    [3, 0, 0, 2]
    [2, 3, 0, 0]
    [2, 2, 1, 0]
    [2, 2, 0, 1]
    [2, 1, 2, 0]
    [2, 1, 0, 2]
    [2, 0, 3, 0]
    [2, 0, 2, 1]
    [2, 0, 1, 2]
    [2, 0, 0, 3]
    [1, 2, 2, 0]
    [1, 2, 0, 2]
    [1, 0, 2, 2]
    [0, 3, 2, 0]
    [0, 3, 0, 2]
    [0, 2, 3, 0]
    [0, 2, 2, 1]
    [0, 2, 1, 2]
    [0, 2, 0, 3]
    [0, 1, 2, 2]
    [0, 0, 3, 2]
    [0, 0, 2, 3]



    I'm having difficulty finding a general formula for this problem. Any help would be appreciated.






    combinatorics







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      0



      $begingroup$


      Suppose I have n objects distributed to k bins and I want to know how many multisets satisfy the following condition: exactly j bins have at least m objects, where j $leq$ k,m $leq$ n, and j*m $leq$ n. For example, if n = 5, k=4, j=2 and m = 2, there are 24 subsets:



      [3, 2, 0, 0]
      [3, 0, 2, 0]
      [3, 0, 0, 2]
      [2, 3, 0, 0]
      [2, 2, 1, 0]
      [2, 2, 0, 1]
      [2, 1, 2, 0]
      [2, 1, 0, 2]
      [2, 0, 3, 0]
      [2, 0, 2, 1]
      [2, 0, 1, 2]
      [2, 0, 0, 3]
      [1, 2, 2, 0]
      [1, 2, 0, 2]
      [1, 0, 2, 2]
      [0, 3, 2, 0]
      [0, 3, 0, 2]
      [0, 2, 3, 0]
      [0, 2, 2, 1]
      [0, 2, 1, 2]
      [0, 2, 0, 3]
      [0, 1, 2, 2]
      [0, 0, 3, 2]
      [0, 0, 2, 3]



      I'm having difficulty finding a general formula for this problem. Any help would be appreciated.






      combinatorics







      share|cite|improve this question









      $endgroup$




      Suppose I have n objects distributed to k bins and I want to know how many multisets satisfy the following condition: exactly j bins have at least m objects, where j $leq$ k,m $leq$ n, and j*m $leq$ n. For example, if n = 5, k=4, j=2 and m = 2, there are 24 subsets:



      [3, 2, 0, 0]
      [3, 0, 2, 0]
      [3, 0, 0, 2]
      [2, 3, 0, 0]
      [2, 2, 1, 0]
      [2, 2, 0, 1]
      [2, 1, 2, 0]
      [2, 1, 0, 2]
      [2, 0, 3, 0]
      [2, 0, 2, 1]
      [2, 0, 1, 2]
      [2, 0, 0, 3]
      [1, 2, 2, 0]
      [1, 2, 0, 2]
      [1, 0, 2, 2]
      [0, 3, 2, 0]
      [0, 3, 0, 2]
      [0, 2, 3, 0]
      [0, 2, 2, 1]
      [0, 2, 1, 2]
      [0, 2, 0, 3]
      [0, 1, 2, 2]
      [0, 0, 3, 2]
      [0, 0, 2, 3]



      I'm having difficulty finding a general formula for this problem. Any help would be appreciated.






      combinatorics




      combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 17:32









      Christopher FisherChristopher Fisher

      32




      32






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          Choose which of the $j$ bins have at least $m$ objects in $binom{k}j$ ways. Let $x_1,x_2,dots,x_j$ be number of objects minus $j$ in these $j$ bins. Let $y_1,y_2,dots,y_{k-j}$ be the number of object in the other bins. Then
          $$
          begin{align}
          x_1+dots+x_j+y_1+dots+y_{k-j}&=n-jm,\
          0 &le x_i,\
          0&le y_i<m
          end{align}
          $$
          If we ignored the condition $y_i<m$, then we would have several non negative integers summing to $n-jm$; the number of ways to do this is $binom{n-jm+k-1}{k-1}$, using stars and bars. In order to account for $y_i<m$, we use inclusion exclusion. Namely, subtract out the bad solutions where $y_ige m$ for each $i$, then add back in the doubly subtracted solutions where $y_ige m$ and $y_jge m$, then subtract out the triply bad solutions, etc. The result is
          $$
          binom{k}jsum_{ell = 0}^{k-j}binom{k-j}{ell}(-1)^ell binom{n-(j+ell)m+k-1}{k-1}
          $$
          Here, I am using the convention that $binom{n}k=0$ when $n<k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
            $endgroup$
            – Christopher Fisher
            Jan 7 at 20:40












          • $begingroup$
            If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
            $endgroup$
            – Mike Earnest
            Jan 7 at 21:10










          • $begingroup$
            You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
            $endgroup$
            – Christopher Fisher
            Jan 7 at 21:23












          • $begingroup$
            The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
            $endgroup$
            – Mike Earnest
            Jan 7 at 22:27











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          1 Answer
          1






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          0












          $begingroup$

          Choose which of the $j$ bins have at least $m$ objects in $binom{k}j$ ways. Let $x_1,x_2,dots,x_j$ be number of objects minus $j$ in these $j$ bins. Let $y_1,y_2,dots,y_{k-j}$ be the number of object in the other bins. Then
          $$
          begin{align}
          x_1+dots+x_j+y_1+dots+y_{k-j}&=n-jm,\
          0 &le x_i,\
          0&le y_i<m
          end{align}
          $$
          If we ignored the condition $y_i<m$, then we would have several non negative integers summing to $n-jm$; the number of ways to do this is $binom{n-jm+k-1}{k-1}$, using stars and bars. In order to account for $y_i<m$, we use inclusion exclusion. Namely, subtract out the bad solutions where $y_ige m$ for each $i$, then add back in the doubly subtracted solutions where $y_ige m$ and $y_jge m$, then subtract out the triply bad solutions, etc. The result is
          $$
          binom{k}jsum_{ell = 0}^{k-j}binom{k-j}{ell}(-1)^ell binom{n-(j+ell)m+k-1}{k-1}
          $$
          Here, I am using the convention that $binom{n}k=0$ when $n<k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
            $endgroup$
            – Christopher Fisher
            Jan 7 at 20:40












          • $begingroup$
            If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
            $endgroup$
            – Mike Earnest
            Jan 7 at 21:10










          • $begingroup$
            You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
            $endgroup$
            – Christopher Fisher
            Jan 7 at 21:23












          • $begingroup$
            The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
            $endgroup$
            – Mike Earnest
            Jan 7 at 22:27
















          0












          $begingroup$

          Choose which of the $j$ bins have at least $m$ objects in $binom{k}j$ ways. Let $x_1,x_2,dots,x_j$ be number of objects minus $j$ in these $j$ bins. Let $y_1,y_2,dots,y_{k-j}$ be the number of object in the other bins. Then
          $$
          begin{align}
          x_1+dots+x_j+y_1+dots+y_{k-j}&=n-jm,\
          0 &le x_i,\
          0&le y_i<m
          end{align}
          $$
          If we ignored the condition $y_i<m$, then we would have several non negative integers summing to $n-jm$; the number of ways to do this is $binom{n-jm+k-1}{k-1}$, using stars and bars. In order to account for $y_i<m$, we use inclusion exclusion. Namely, subtract out the bad solutions where $y_ige m$ for each $i$, then add back in the doubly subtracted solutions where $y_ige m$ and $y_jge m$, then subtract out the triply bad solutions, etc. The result is
          $$
          binom{k}jsum_{ell = 0}^{k-j}binom{k-j}{ell}(-1)^ell binom{n-(j+ell)m+k-1}{k-1}
          $$
          Here, I am using the convention that $binom{n}k=0$ when $n<k$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
            $endgroup$
            – Christopher Fisher
            Jan 7 at 20:40












          • $begingroup$
            If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
            $endgroup$
            – Mike Earnest
            Jan 7 at 21:10










          • $begingroup$
            You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
            $endgroup$
            – Christopher Fisher
            Jan 7 at 21:23












          • $begingroup$
            The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
            $endgroup$
            – Mike Earnest
            Jan 7 at 22:27














          0












          0








          0





          $begingroup$

          Choose which of the $j$ bins have at least $m$ objects in $binom{k}j$ ways. Let $x_1,x_2,dots,x_j$ be number of objects minus $j$ in these $j$ bins. Let $y_1,y_2,dots,y_{k-j}$ be the number of object in the other bins. Then
          $$
          begin{align}
          x_1+dots+x_j+y_1+dots+y_{k-j}&=n-jm,\
          0 &le x_i,\
          0&le y_i<m
          end{align}
          $$
          If we ignored the condition $y_i<m$, then we would have several non negative integers summing to $n-jm$; the number of ways to do this is $binom{n-jm+k-1}{k-1}$, using stars and bars. In order to account for $y_i<m$, we use inclusion exclusion. Namely, subtract out the bad solutions where $y_ige m$ for each $i$, then add back in the doubly subtracted solutions where $y_ige m$ and $y_jge m$, then subtract out the triply bad solutions, etc. The result is
          $$
          binom{k}jsum_{ell = 0}^{k-j}binom{k-j}{ell}(-1)^ell binom{n-(j+ell)m+k-1}{k-1}
          $$
          Here, I am using the convention that $binom{n}k=0$ when $n<k$.






          share|cite|improve this answer









          $endgroup$



          Choose which of the $j$ bins have at least $m$ objects in $binom{k}j$ ways. Let $x_1,x_2,dots,x_j$ be number of objects minus $j$ in these $j$ bins. Let $y_1,y_2,dots,y_{k-j}$ be the number of object in the other bins. Then
          $$
          begin{align}
          x_1+dots+x_j+y_1+dots+y_{k-j}&=n-jm,\
          0 &le x_i,\
          0&le y_i<m
          end{align}
          $$
          If we ignored the condition $y_i<m$, then we would have several non negative integers summing to $n-jm$; the number of ways to do this is $binom{n-jm+k-1}{k-1}$, using stars and bars. In order to account for $y_i<m$, we use inclusion exclusion. Namely, subtract out the bad solutions where $y_ige m$ for each $i$, then add back in the doubly subtracted solutions where $y_ige m$ and $y_jge m$, then subtract out the triply bad solutions, etc. The result is
          $$
          binom{k}jsum_{ell = 0}^{k-j}binom{k-j}{ell}(-1)^ell binom{n-(j+ell)m+k-1}{k-1}
          $$
          Here, I am using the convention that $binom{n}k=0$ when $n<k$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 18:46









          Mike EarnestMike Earnest

          23.3k12051




          23.3k12051












          • $begingroup$
            Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
            $endgroup$
            – Christopher Fisher
            Jan 7 at 20:40












          • $begingroup$
            If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
            $endgroup$
            – Mike Earnest
            Jan 7 at 21:10










          • $begingroup$
            You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
            $endgroup$
            – Christopher Fisher
            Jan 7 at 21:23












          • $begingroup$
            The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
            $endgroup$
            – Mike Earnest
            Jan 7 at 22:27


















          • $begingroup$
            Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
            $endgroup$
            – Christopher Fisher
            Jan 7 at 20:40












          • $begingroup$
            If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
            $endgroup$
            – Mike Earnest
            Jan 7 at 21:10










          • $begingroup$
            You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
            $endgroup$
            – Christopher Fisher
            Jan 7 at 21:23












          • $begingroup$
            The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
            $endgroup$
            – Mike Earnest
            Jan 7 at 22:27
















          $begingroup$
          Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
          $endgroup$
          – Christopher Fisher
          Jan 7 at 20:40






          $begingroup$
          Thank you for your help. I believe I identified a case where it fails: n=4,k=3,m=4,j=1. I think it should equal ${k choose j}$ = 3. I suspect the inclusion-exclusion method might be failing because something is overcounted in certain cases.
          $endgroup$
          – Christopher Fisher
          Jan 7 at 20:40














          $begingroup$
          If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
          $endgroup$
          – Mike Earnest
          Jan 7 at 21:10




          $begingroup$
          If you apply my formula in your comment, you should get $$binom{3}1left(binom{2}0binom{2}2-binom{2}1binom{-2}{2}+binom{2}2binom{-6}2right)=3(1-0+0)=3,$$ which is correct.
          $endgroup$
          – Mike Earnest
          Jan 7 at 21:10












          $begingroup$
          You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
          $endgroup$
          – Christopher Fisher
          Jan 7 at 21:23






          $begingroup$
          You are right. Sorry for the confusion. The program I am using yielded ${-2 choose 1}$ = -2, which was unexpected. Thanks again!
          $endgroup$
          – Christopher Fisher
          Jan 7 at 21:23














          $begingroup$
          The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
          $endgroup$
          – Mike Earnest
          Jan 7 at 22:27




          $begingroup$
          The confusion is my fault. Most mathematicians agree that $binom{-2}1=-2$; it just makes my formula more convenient to write if you assume $binom{-2}1=0$.
          $endgroup$
          – Mike Earnest
          Jan 7 at 22:27


















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          張江高科駅

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          張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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