Mixing
Summation of: (binomial coefficients * Stirling numbers of the second kind)
$begingroup$
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
edited Jan 16 at 19:59
Bernard
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 19:59
Bernard
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 19:59
Bernard
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
$endgroup$
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
edited Jan 16 at 19:59
Bernard
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
edited Jan 16 at 19:59
Bernard
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
edited Jan 16 at 19:59
Bernard
121k740116
edited Jan 16 at 19:59
Bernard
121k740116
edited Jan 16 at 19:59
Bernard
121k740116
121k740116
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
asked Jan 16 at 19:53
Novice GeekNovice Geek
3811
3811
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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*}
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
add a comment |
$begingroup$
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 $K$.
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
$endgroup$
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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*}
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
add a comment |
$begingroup$
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*}
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
add a comment |
$begingroup$
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*}
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
$endgroup$
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*}
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
answered Jan 16 at 22:03
Donald SplutterwitDonald Splutterwit
22.8k21445
22.8k21445
add a comment |
add a comment |
$begingroup$
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 $K$.
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
$endgroup$
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
add a comment |
$begingroup$
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 $K$.
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
$endgroup$
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
add a comment |
$begingroup$
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 $K$.
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
$endgroup$
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 $K$.
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
edited Jan 16 at 23:43
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
answered Jan 16 at 23:15
Mike EarnestMike Earnest
23.5k12051
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
add a comment |
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
add a comment |
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