Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find a closed form for $a_n:=\sum_{k=0}^{n}\binom{n}{k}(n-k)^n(-1)^k$ using generating functions.

share|improve this question
4  
With that tone of question, it might help if you showed us what you have tried. –  Henry Jun 27 '12 at 20:17
    
@akotronis: (+1) for the interesting question. –  user26872 Jun 28 '12 at 6:24
add comment

1 Answer

up vote 3 down vote accepted

We have $$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k \binom{n}{k}(n-k)^n &=& \sum_{k=0}^{n}(-1)^k \binom{n}{n-k}(n-k)^n \\ &=& \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} k^n \\ &=& \left.\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} (x D)^n x^k\right|_{x=1} \\ &=& \left.(x D)^n \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} x^k\right|_{x=1} \\ &=& \left.(x D)^n (x-1)^n\right|_{x=1}, \end{eqnarray*}$$ where $D = \partial/\partial x$. But $$(x D)^n = x^n D^n + (\mathrm{const}) x^{n-1}D^{n-1} + \ldots.$$ and $D^k(x-1)^n|_{x=1} = 0$ unless $k\ge n$. Therefore, $\left.(x D)^n (x-1)^n\right|_{x=1} = D^n(x-1)^n|_{x=1} = n!,$ and so $$\begin{equation*} \sum_{k=0}^{n}(-1)^k \binom{n}{k}(n-k)^n = n!.\tag{1} \end{equation*}$$


The argument above immediately implies that $$\sum_{k=0}^{n}(-1)^k \binom{n}{k}(n-k)^m = 0$$ if $m\in\mathbb{N}$ and $m<n$. It also gives us a method to calculate the sum for $m>n$. Sums of this type are related to the Stirling numbers of the second kind, $$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k \binom{n}{k}(n-k)^m &=& \sum_{k=0}^{n}(-1)^{n-k} \binom{n}{k}k^m \\ &=& n! \left\{m\atop n\right\}. \end{eqnarray*}$$ The operator $(x D)^n$ and it's connection to the Stirling numbers has been discussed here.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.