Distribute distinct objects in identical boxes












1












$begingroup$


Number of ways to distribute $6$ distinct objects to $3$ identical boxes such that each box should have atleast one?



$mathbf {Is there any standard formula for these sums}$, as we have for $langle identical object, distinct boxrangle$ pair as $$N+R-1choose R-1$$






combinatorics stirling-numbers balls-in-bins







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    $begingroup$


    Number of ways to distribute $6$ distinct objects to $3$ identical boxes such that each box should have atleast one?



    $mathbf {Is there any standard formula for these sums}$, as we have for $langle identical object, distinct boxrangle$ pair as $$N+R-1choose R-1$$






    combinatorics stirling-numbers balls-in-bins







    share|cite|improve this question











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      1












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      1


      2



      $begingroup$


      Number of ways to distribute $6$ distinct objects to $3$ identical boxes such that each box should have atleast one?



      $mathbf {Is there any standard formula for these sums}$, as we have for $langle identical object, distinct boxrangle$ pair as $$N+R-1choose R-1$$






      combinatorics stirling-numbers balls-in-bins







      share|cite|improve this question











      $endgroup$




      Number of ways to distribute $6$ distinct objects to $3$ identical boxes such that each box should have atleast one?



      $mathbf {Is there any standard formula for these sums}$, as we have for $langle identical object, distinct boxrangle$ pair as $$N+R-1choose R-1$$






      combinatorics stirling-numbers balls-in-bins




      combinatorics stirling-numbers balls-in-bins






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Oct 5 '13 at 19:25









      dibyendu

      356318




      356318










      asked Sep 4 '12 at 6:14









      AizenAizen

      14917




      14917






















          1 Answer
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          4












          $begingroup$

          These numbers are the Stirling numbers of the second kind; the specific one that you mention is denoted by $left{6atop 3right}$ or $S(6,3)$. These numbers satisfy a fairly simple recurrence relation that is reminiscent of the relation $dbinom{n+1}k=dbinom{n}k+dbinom{n}{k-1}$ satisfied by the binomial coefficients:



          $$left{n+1atop kright}=kleft{natop kright}+left{natop k-1right};,$$



          with initial conditions $left{0atop 0right}=1$ and $left{natop 0right}=left{0atop nright}=0$ for $n>0$.




          Added: This recurrence isn’t hard to prove. Consider the ways of partitioning the integers $1,dots,n+1$ into $k$ non-empty sets. We can start with a partition of ${1,dots,n}$ into $k-1$ non-empty sets and make ${n+1}$ the $k$-th set; there are $left{natop k-1right}$ ways to do this. Or we can partition ${1,dots,n}$ into $k$ non-empty sets and add $n+1$ to one of those $k$ sets; there are $kleft{natop kright}$ ways to do this. The total number of possibilities is therefore $kleft{natop kright}+left{natop k-1right}$.




          There is an explicit formula for $left{natop kright}$, but it’s rather ugly:



          $$left{natop kright}=frac1{k!}sum_{i=0}^k(-1)^{k-i}binom{k}ii^n;.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If its asked not at least 1 but any numbers then or for at least 2 ??
            $endgroup$
            – Aizen
            Sep 4 '12 at 6:38










          • $begingroup$
            @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:24










          • $begingroup$
            But inclusion-exclusion is valid when both are distinct.
            $endgroup$
            – Aizen
            Sep 4 '12 at 7:37










          • $begingroup$
            @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:40











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          1 Answer
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          1 Answer
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          active

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          4












          $begingroup$

          These numbers are the Stirling numbers of the second kind; the specific one that you mention is denoted by $left{6atop 3right}$ or $S(6,3)$. These numbers satisfy a fairly simple recurrence relation that is reminiscent of the relation $dbinom{n+1}k=dbinom{n}k+dbinom{n}{k-1}$ satisfied by the binomial coefficients:



          $$left{n+1atop kright}=kleft{natop kright}+left{natop k-1right};,$$



          with initial conditions $left{0atop 0right}=1$ and $left{natop 0right}=left{0atop nright}=0$ for $n>0$.




          Added: This recurrence isn’t hard to prove. Consider the ways of partitioning the integers $1,dots,n+1$ into $k$ non-empty sets. We can start with a partition of ${1,dots,n}$ into $k-1$ non-empty sets and make ${n+1}$ the $k$-th set; there are $left{natop k-1right}$ ways to do this. Or we can partition ${1,dots,n}$ into $k$ non-empty sets and add $n+1$ to one of those $k$ sets; there are $kleft{natop kright}$ ways to do this. The total number of possibilities is therefore $kleft{natop kright}+left{natop k-1right}$.




          There is an explicit formula for $left{natop kright}$, but it’s rather ugly:



          $$left{natop kright}=frac1{k!}sum_{i=0}^k(-1)^{k-i}binom{k}ii^n;.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If its asked not at least 1 but any numbers then or for at least 2 ??
            $endgroup$
            – Aizen
            Sep 4 '12 at 6:38










          • $begingroup$
            @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:24










          • $begingroup$
            But inclusion-exclusion is valid when both are distinct.
            $endgroup$
            – Aizen
            Sep 4 '12 at 7:37










          • $begingroup$
            @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:40
















          4












          $begingroup$

          These numbers are the Stirling numbers of the second kind; the specific one that you mention is denoted by $left{6atop 3right}$ or $S(6,3)$. These numbers satisfy a fairly simple recurrence relation that is reminiscent of the relation $dbinom{n+1}k=dbinom{n}k+dbinom{n}{k-1}$ satisfied by the binomial coefficients:



          $$left{n+1atop kright}=kleft{natop kright}+left{natop k-1right};,$$



          with initial conditions $left{0atop 0right}=1$ and $left{natop 0right}=left{0atop nright}=0$ for $n>0$.




          Added: This recurrence isn’t hard to prove. Consider the ways of partitioning the integers $1,dots,n+1$ into $k$ non-empty sets. We can start with a partition of ${1,dots,n}$ into $k-1$ non-empty sets and make ${n+1}$ the $k$-th set; there are $left{natop k-1right}$ ways to do this. Or we can partition ${1,dots,n}$ into $k$ non-empty sets and add $n+1$ to one of those $k$ sets; there are $kleft{natop kright}$ ways to do this. The total number of possibilities is therefore $kleft{natop kright}+left{natop k-1right}$.




          There is an explicit formula for $left{natop kright}$, but it’s rather ugly:



          $$left{natop kright}=frac1{k!}sum_{i=0}^k(-1)^{k-i}binom{k}ii^n;.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If its asked not at least 1 but any numbers then or for at least 2 ??
            $endgroup$
            – Aizen
            Sep 4 '12 at 6:38










          • $begingroup$
            @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:24










          • $begingroup$
            But inclusion-exclusion is valid when both are distinct.
            $endgroup$
            – Aizen
            Sep 4 '12 at 7:37










          • $begingroup$
            @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:40














          4












          4








          4





          $begingroup$

          These numbers are the Stirling numbers of the second kind; the specific one that you mention is denoted by $left{6atop 3right}$ or $S(6,3)$. These numbers satisfy a fairly simple recurrence relation that is reminiscent of the relation $dbinom{n+1}k=dbinom{n}k+dbinom{n}{k-1}$ satisfied by the binomial coefficients:



          $$left{n+1atop kright}=kleft{natop kright}+left{natop k-1right};,$$



          with initial conditions $left{0atop 0right}=1$ and $left{natop 0right}=left{0atop nright}=0$ for $n>0$.




          Added: This recurrence isn’t hard to prove. Consider the ways of partitioning the integers $1,dots,n+1$ into $k$ non-empty sets. We can start with a partition of ${1,dots,n}$ into $k-1$ non-empty sets and make ${n+1}$ the $k$-th set; there are $left{natop k-1right}$ ways to do this. Or we can partition ${1,dots,n}$ into $k$ non-empty sets and add $n+1$ to one of those $k$ sets; there are $kleft{natop kright}$ ways to do this. The total number of possibilities is therefore $kleft{natop kright}+left{natop k-1right}$.




          There is an explicit formula for $left{natop kright}$, but it’s rather ugly:



          $$left{natop kright}=frac1{k!}sum_{i=0}^k(-1)^{k-i}binom{k}ii^n;.$$






          share|cite|improve this answer











          $endgroup$



          These numbers are the Stirling numbers of the second kind; the specific one that you mention is denoted by $left{6atop 3right}$ or $S(6,3)$. These numbers satisfy a fairly simple recurrence relation that is reminiscent of the relation $dbinom{n+1}k=dbinom{n}k+dbinom{n}{k-1}$ satisfied by the binomial coefficients:



          $$left{n+1atop kright}=kleft{natop kright}+left{natop k-1right};,$$



          with initial conditions $left{0atop 0right}=1$ and $left{natop 0right}=left{0atop nright}=0$ for $n>0$.




          Added: This recurrence isn’t hard to prove. Consider the ways of partitioning the integers $1,dots,n+1$ into $k$ non-empty sets. We can start with a partition of ${1,dots,n}$ into $k-1$ non-empty sets and make ${n+1}$ the $k$-th set; there are $left{natop k-1right}$ ways to do this. Or we can partition ${1,dots,n}$ into $k$ non-empty sets and add $n+1$ to one of those $k$ sets; there are $kleft{natop kright}$ ways to do this. The total number of possibilities is therefore $kleft{natop kright}+left{natop k-1right}$.




          There is an explicit formula for $left{natop kright}$, but it’s rather ugly:



          $$left{natop kright}=frac1{k!}sum_{i=0}^k(-1)^{k-i}binom{k}ii^n;.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Sep 4 '12 at 6:31

























          answered Sep 4 '12 at 6:23









          Brian M. ScottBrian M. Scott

          459k38513916




          459k38513916












          • $begingroup$
            If its asked not at least 1 but any numbers then or for at least 2 ??
            $endgroup$
            – Aizen
            Sep 4 '12 at 6:38










          • $begingroup$
            @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:24










          • $begingroup$
            But inclusion-exclusion is valid when both are distinct.
            $endgroup$
            – Aizen
            Sep 4 '12 at 7:37










          • $begingroup$
            @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:40


















          • $begingroup$
            If its asked not at least 1 but any numbers then or for at least 2 ??
            $endgroup$
            – Aizen
            Sep 4 '12 at 6:38










          • $begingroup$
            @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:24










          • $begingroup$
            But inclusion-exclusion is valid when both are distinct.
            $endgroup$
            – Aizen
            Sep 4 '12 at 7:37










          • $begingroup$
            @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
            $endgroup$
            – Brian M. Scott
            Sep 4 '12 at 7:40
















          $begingroup$
          If its asked not at least 1 but any numbers then or for at least 2 ??
          $endgroup$
          – Aizen
          Sep 4 '12 at 6:38




          $begingroup$
          If its asked not at least 1 but any numbers then or for at least 2 ??
          $endgroup$
          – Aizen
          Sep 4 '12 at 6:38












          $begingroup$
          @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
          $endgroup$
          – Brian M. Scott
          Sep 4 '12 at 7:24




          $begingroup$
          @Aizen: If each box must contain at least two objects, an inclusion-exclusion argument shows that the number of arrangements is $$sum_{i=1}^{k-1}(-1)^{k-1}binom{n}ileft{n-iatop k-iright};.$$ It’s entirely possible that this can be simplified, but I don’t immediately see anything nice. After the ‘at least $2$’ case it looks to be getting really messy; I’d probably try to work with generating functions at that point.
          $endgroup$
          – Brian M. Scott
          Sep 4 '12 at 7:24












          $begingroup$
          But inclusion-exclusion is valid when both are distinct.
          $endgroup$
          – Aizen
          Sep 4 '12 at 7:37




          $begingroup$
          But inclusion-exclusion is valid when both are distinct.
          $endgroup$
          – Aizen
          Sep 4 '12 at 7:37












          $begingroup$
          @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
          $endgroup$
          – Brian M. Scott
          Sep 4 '12 at 7:40




          $begingroup$
          @Aizen: I don’t understand what you’re trying to say. Inclusion-exclusion is a general technique that is applicable in a wide variety of situations.
          $endgroup$
          – Brian M. Scott
          Sep 4 '12 at 7:40


















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