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When $k \lt n$, what is the value of the sum $$\sum\limits_{j=0}^n {n \choose j}(-1)^j (n-j)^k.$$ Explain combinatorially.

Any ideas on where to start?

Martin Sleziak
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user67241
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  • Is your formula correct? As written now, the factor $\binom nk$ is independent of $j$ and can just be factored out of the sum. – Greg Martin Mar 18 '13 at 06:21
  • Sorry, fixed it – user67241 Mar 18 '13 at 06:30
  • Checked again. My answer is okay. Mine is also consistent with Stirling number. Derived the combinatorial meaning just by looking at the sum. But, the other answer reminded me of Stirling number. – Sungjin Kim Mar 18 '13 at 07:41

3 Answers3

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This is inclusion-exclusion principle.

Consider the functions from a set $S$ with $k$ elements to a set $T$ with $n$ elements. Let $A_j$ be the functions whose range excludes a fixed subset of $j$ elements. The sum is $$ \sum_{j=0}^n (-1)^j \binom{n}{j}|A_j|.$$

Then the sum represents the number of all onto functions from $S$ to $T$. (Inclusion-exclusion principle)

Since $k<n$, this number is clearly zero.

Also, there is an analytic proof of this. Consider the function $$f(x)=\sum_{j=0}^n\binom{n}{j}(-1)^je^{(n-j)x} = (e^x -1)^n$$ Then the sum represents, the $k$-th derivative of $f$ at 0, say $f^{(k)}(0)$. Since $k<n$, this also gives zero.

Sungjin Kim
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The sum vanishes. It counts the ordered partitions of $k$ elements into $n$ nonempty subsets, and there are none for $k\lt n$. This is $\displaystyle n!\left\{k\atop n\right\}$, where $\displaystyle\left\{k\atop n\right\}$ is a Stirling number of the second kind.

joriki
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  • It should be partitioning the set of $k$ elements, with $n$ nonempty parts. – Sungjin Kim Mar 18 '13 at 07:50
  • @i707107: I think that's what it says; you may be referring to the initial version that was very briefly visible, where I had it the wrong way around. – joriki Mar 18 '13 at 08:43
  • I am just talking about the second sentence. "$n$-tuples of non-empty subsets of $k$ elements" is not a partition. – Sungjin Kim Mar 18 '13 at 09:03
  • @i707107: I see, thanks, I thought you meant I'd switched $n$ and $k$ (which I originally had). You're right, I failed to state that the subsets form a partition. I wanted to phrase it in terms of $n$-tuples to emphasize that the partitions are ordered, but I guess "ordered partitions" is clear enough. – joriki Mar 18 '13 at 12:42
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As I explained in this other answer, you can recognise the alternating sum along row $n$ of Pascal's triangle, with the index (here $j$) used as a shift in the remaining expression (here $(n-j)^k$), as an application of the $n$-th power of the finite difference operator $-\Delta:f\mapsto(x\mapsto f(x)-f(x+1))$, and this operator decreases the degree of polynomial functions by $1$, making all those of degree less than $n$ identically zero. Here one has, from the fact that $(-\Delta)^n(f)=0$ when $f(x)=x^k$, that $$ \sum_{i=0}^n(-1)^i\binom ni (x+i)^k = 0\qquad\text{for all $x\in\mathbf C$, if $k<n$.} $$ Now taking $x=-n$ it easily follows that the summation in the question vanishes.

As for a combinatorial interpretation for the summation, one that comes to mind is a signed summation over all ways to forbid a subset of $j$ out of $n$ elements, and choose a map from a $k$-set to the remaining elements (an element may be not-forbidden but still not in the image of the map), with the sign being given by the partity of the number of forbidden elements. Since $k<n$, no map from a $k$-set to a $n$-set can be surjective; one can take the first element not in the image and flip its "forbidded status", and this gives a sign-reversing involution on the set, so the signed sum is $0$.

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Marc van Leeuwen
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