Summation of: (binomial coefficients * Stirling numbers of the second kind)












3












$begingroup$


Problem: Simplify the following equation:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k!
end{equation}



A solution: I am aware of the following relation:
begin{equation}
sumlimits_{t=1}^n t^n = sumlimits_{k=1}^n dbinom{n+1}{k+1} begin{Bmatrix} n\ k end{Bmatrix} k!
end{equation}



Now, I am struggling to get to a somewhat similar (i.e., clean) expression for the given equation.



Thanks for your help!






combinatorics binomial-coefficients stirling-numbers







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$endgroup$

















    3












    $begingroup$


    Problem: Simplify the following equation:
    begin{equation}
    sumlimits_{k=1}^n dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k!
    end{equation}



    A solution: I am aware of the following relation:
    begin{equation}
    sumlimits_{t=1}^n t^n = sumlimits_{k=1}^n dbinom{n+1}{k+1} begin{Bmatrix} n\ k end{Bmatrix} k!
    end{equation}



    Now, I am struggling to get to a somewhat similar (i.e., clean) expression for the given equation.



    Thanks for your help!






    combinatorics binomial-coefficients stirling-numbers







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Problem: Simplify the following equation:
      begin{equation}
      sumlimits_{k=1}^n dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k!
      end{equation}



      A solution: I am aware of the following relation:
      begin{equation}
      sumlimits_{t=1}^n t^n = sumlimits_{k=1}^n dbinom{n+1}{k+1} begin{Bmatrix} n\ k end{Bmatrix} k!
      end{equation}



      Now, I am struggling to get to a somewhat similar (i.e., clean) expression for the given equation.



      Thanks for your help!






      combinatorics binomial-coefficients stirling-numbers







      share|cite|improve this question











      $endgroup$




      Problem: Simplify the following equation:
      begin{equation}
      sumlimits_{k=1}^n dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k!
      end{equation}



      A solution: I am aware of the following relation:
      begin{equation}
      sumlimits_{t=1}^n t^n = sumlimits_{k=1}^n dbinom{n+1}{k+1} begin{Bmatrix} n\ k end{Bmatrix} k!
      end{equation}



      Now, I am struggling to get to a somewhat similar (i.e., clean) expression for the given equation.



      Thanks for your help!






      combinatorics binomial-coefficients stirling-numbers




      combinatorics binomial-coefficients stirling-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 19:59









      Bernard

      121k740116




      121k740116










      asked Jan 16 at 19:53









      Novice GeekNovice Geek

      3811




      3811






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.



          We recall the well known expression for the Stirling numbers of the second kind
          begin{eqnarray*}
          begin{Bmatrix} n\ k end{Bmatrix} k! = n! [x^n]:(e^x-1)^k.
          end{eqnarray*}
          So your sum can be rewritten as
          begin{eqnarray*}
          sum_{k=1}^{n} dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k! &=& n! [x^n]: sum_{k=1}^{n} dbinom{n}{k}(e^x-1)^k \
          &=& n! [x^n]: (1+e^x-1)^n -1 \
          &=& n! [x^n]: e^{nx} -1 =color{red}{n^n}.
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            The answer is $n^n$, as mentioned in Donald's answer, and here is a combinatorial proof.



            Letting $[n]={1,2,dots,n}$, a function from $f:[n]to [n]$ is specified by




            • a subset $K$ of $[n]$ of size $k$ , for some $1le kle n$, equal to the range of $f$,

            • a partition of the domain $[n]$ into $k$ parts, and

            • an ordering of those $k$ parts.


            The interpretation is that $f(x) = y$ when $x$ is in part number $i$, and $y$ is the $i^{th}$ smallest element of $K$.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
              $endgroup$
              – Alexander Burstein
              Jan 17 at 3:47













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.



            We recall the well known expression for the Stirling numbers of the second kind
            begin{eqnarray*}
            begin{Bmatrix} n\ k end{Bmatrix} k! = n! [x^n]:(e^x-1)^k.
            end{eqnarray*}
            So your sum can be rewritten as
            begin{eqnarray*}
            sum_{k=1}^{n} dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k! &=& n! [x^n]: sum_{k=1}^{n} dbinom{n}{k}(e^x-1)^k \
            &=& n! [x^n]: (1+e^x-1)^n -1 \
            &=& n! [x^n]: e^{nx} -1 =color{red}{n^n}.
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.



              We recall the well known expression for the Stirling numbers of the second kind
              begin{eqnarray*}
              begin{Bmatrix} n\ k end{Bmatrix} k! = n! [x^n]:(e^x-1)^k.
              end{eqnarray*}
              So your sum can be rewritten as
              begin{eqnarray*}
              sum_{k=1}^{n} dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k! &=& n! [x^n]: sum_{k=1}^{n} dbinom{n}{k}(e^x-1)^k \
              &=& n! [x^n]: (1+e^x-1)^n -1 \
              &=& n! [x^n]: e^{nx} -1 =color{red}{n^n}.
              end{eqnarray*}






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.



                We recall the well known expression for the Stirling numbers of the second kind
                begin{eqnarray*}
                begin{Bmatrix} n\ k end{Bmatrix} k! = n! [x^n]:(e^x-1)^k.
                end{eqnarray*}
                So your sum can be rewritten as
                begin{eqnarray*}
                sum_{k=1}^{n} dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k! &=& n! [x^n]: sum_{k=1}^{n} dbinom{n}{k}(e^x-1)^k \
                &=& n! [x^n]: (1+e^x-1)^n -1 \
                &=& n! [x^n]: e^{nx} -1 =color{red}{n^n}.
                end{eqnarray*}






                share|cite|improve this answer









                $endgroup$



                Using the $[x^k]:f(x)$ to represent the coeficient of $x^k$ in the power series for $f(x)$.



                We recall the well known expression for the Stirling numbers of the second kind
                begin{eqnarray*}
                begin{Bmatrix} n\ k end{Bmatrix} k! = n! [x^n]:(e^x-1)^k.
                end{eqnarray*}
                So your sum can be rewritten as
                begin{eqnarray*}
                sum_{k=1}^{n} dbinom{n}{k} begin{Bmatrix} n\ k end{Bmatrix} k! &=& n! [x^n]: sum_{k=1}^{n} dbinom{n}{k}(e^x-1)^k \
                &=& n! [x^n]: (1+e^x-1)^n -1 \
                &=& n! [x^n]: e^{nx} -1 =color{red}{n^n}.
                end{eqnarray*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 22:03









                Donald SplutterwitDonald Splutterwit

                22.8k21445




                22.8k21445























                    3












                    $begingroup$

                    The answer is $n^n$, as mentioned in Donald's answer, and here is a combinatorial proof.



                    Letting $[n]={1,2,dots,n}$, a function from $f:[n]to [n]$ is specified by




                    • a subset $K$ of $[n]$ of size $k$ , for some $1le kle n$, equal to the range of $f$,

                    • a partition of the domain $[n]$ into $k$ parts, and

                    • an ordering of those $k$ parts.


                    The interpretation is that $f(x) = y$ when $x$ is in part number $i$, and $y$ is the $i^{th}$ smallest element of $K$.






                    share|cite|improve this answer











                    $endgroup$









                    • 2




                      $begingroup$
                      That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                      $endgroup$
                      – Alexander Burstein
                      Jan 17 at 3:47


















                    3












                    $begingroup$

                    The answer is $n^n$, as mentioned in Donald's answer, and here is a combinatorial proof.



                    Letting $[n]={1,2,dots,n}$, a function from $f:[n]to [n]$ is specified by




                    • a subset $K$ of $[n]$ of size $k$ , for some $1le kle n$, equal to the range of $f$,

                    • a partition of the domain $[n]$ into $k$ parts, and

                    • an ordering of those $k$ parts.


                    The interpretation is that $f(x) = y$ when $x$ is in part number $i$, and $y$ is the $i^{th}$ smallest element of $K$.






                    share|cite|improve this answer











                    $endgroup$









                    • 2




                      $begingroup$
                      That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                      $endgroup$
                      – Alexander Burstein
                      Jan 17 at 3:47
















                    3












                    3








                    3





                    $begingroup$

                    The answer is $n^n$, as mentioned in Donald's answer, and here is a combinatorial proof.



                    Letting $[n]={1,2,dots,n}$, a function from $f:[n]to [n]$ is specified by




                    • a subset $K$ of $[n]$ of size $k$ , for some $1le kle n$, equal to the range of $f$,

                    • a partition of the domain $[n]$ into $k$ parts, and

                    • an ordering of those $k$ parts.


                    The interpretation is that $f(x) = y$ when $x$ is in part number $i$, and $y$ is the $i^{th}$ smallest element of $K$.






                    share|cite|improve this answer











                    $endgroup$



                    The answer is $n^n$, as mentioned in Donald's answer, and here is a combinatorial proof.



                    Letting $[n]={1,2,dots,n}$, a function from $f:[n]to [n]$ is specified by




                    • a subset $K$ of $[n]$ of size $k$ , for some $1le kle n$, equal to the range of $f$,

                    • a partition of the domain $[n]$ into $k$ parts, and

                    • an ordering of those $k$ parts.


                    The interpretation is that $f(x) = y$ when $x$ is in part number $i$, and $y$ is the $i^{th}$ smallest element of $K$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 16 at 23:43

























                    answered Jan 16 at 23:15









                    Mike EarnestMike Earnest

                    23.5k12051




                    23.5k12051








                    • 2




                      $begingroup$
                      That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                      $endgroup$
                      – Alexander Burstein
                      Jan 17 at 3:47
















                    • 2




                      $begingroup$
                      That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                      $endgroup$
                      – Alexander Burstein
                      Jan 17 at 3:47










                    2




                    2




                    $begingroup$
                    That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                    $endgroup$
                    – Alexander Burstein
                    Jan 17 at 3:47






                    $begingroup$
                    That's essentially a special case of the fact that any function is a composition of a surjection and an injection. Relating this to the 12-fold way in this case yields the desired formula.
                    $endgroup$
                    – Alexander Burstein
                    Jan 17 at 3:47




















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