Alternating binomial sum of $i^a(n-i)^b$ I would like to know of value (if not, asymptotic growth in $n,a,b$) of the following alternating sum:
$$\sum_{i=0}^n(-1)^i\binom{n}{i}i^a(n-i)^b.$$
Here $a$ and $b$ are natural numbers. When $a+b\leq n$, the above sum is zero. What about $a+b>n$? For instance, I'm especially interested in the case when $a=b=n$. Thanks.
 A: Luckily when $a=b=n$ and $n$ is odd the sum is zero:
$$S_n=\sum_{k=0}^{n} (-1)^k {n \choose k}~ k^n ~ (n-k)^n$$
replace $k$ by $n-k$ (summing backwards), then
$$S_n=\sum_{k=0}^{n} (-1)^k {n \choose n-k}~(n-k)^n ~ k^n=(-1)^n S_n.$$ So when $n$ is odd $S_n=0$
A: Let ${n \brace k}$ denote the $(n,k)^{th}$ Stirling number of the second kind, equal to the number of partitions of a set of size $n$ into $k$ parts. We can show that the sum evaluates to
$$
n!\sum_{k=0}^n(-1)^k{a \brace k}{b \brace n-k}
$$

Let $[n]$ denote $\{1,\dots,n\}$.
The summation $\sum_{i=0}^n \binom{n}{i} (-1)^i i^a(n-i)^b$ enumerates tuples $(I,f,g)$, where

*

*$I\subseteq [n]$,


*$f$ is a function from $[a]$ to $I$,


*$g$ is a function from $[b]$ to $[n]\setminus I$,
and where the contribution of a tuple is $(-1)^{|I|}$. Consider the following sign-reversing involution defined on the set of tuples $(I,f,g)$:

Given $(I,f,g)$, let $m$ be the smallest element of $[n]\setminus (\text{im }f\cup \text{im }g)$. (If $[n]\setminus (\text{im }f\cup \text{im }g)$ is empty, the involution is undefined). Then the involution is $$(I,f,g)\longleftrightarrow (I\oplus \{m\},f,g),$$where $\oplus$ is the symmetric difference of sets. That is, if $m\in I$, remove it, and if $m\notin I$, then add it.

This involution divides the set of tuples $(I,f,g)$ into pairs which cancel out in $\sum_{i=0}^n\binom{n}i (-1)^ii^a(n-i)^b$, so anything in the domain of the involution contributes nothing to the sum. What remains are pairs $(I,f,g)$ where $f:[a]\to I$ is surjective and $g:[b]\to ([n]\setminus I)$ is surjective. The number of surjective functions $f:[a]\to I$ is $|I|!\cdot {a\brace |I|}$. Therefore,
$$
\sum_{i=0}^n\binom{n}i(-1)^ii^a(n-i)^b
=\sum_{I\subseteq [n]} (-1)^{|I|}\cdot|I|!{a\brace |I|}\cdot (n-|I|)!{b\brace n-|I|}\\
\phantom{\sum_{i=0}^n\binom{n}i(-1)^ii^a(n-i)^b}
=n!\sum_{i=0}^n(-1)^i{a\brace i}{b\brace n-i} 
$$

In particular, with $a=b=n$, we get
$$
n!\sum_{i=0}^n (-1)^i {n\brace i}{n\brace n-i}
$$
but I cannot determine the asymptotic of this sequence. It is not even in OEIS...
