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$\sum_{m} \binom {n}{m}S(m,k)S(n-m,l)= \binom{k + l}{l}S(n,k + l)$. How do I give a combinatorial proof of this identity?

My attempt:

LHS: choose m elements out of $n$ total elements then partition $m$ into $k$ and the rest into $l$.

RHS: partition n elements into $k + l$ boxes, then from the boxes choose $l$ elements to be special.

glS
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mac
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  • Did you try writing down the definition of the Stirling numbers into the equation? Maybe take a look at https://math.stackexchange.com/questions/1279874/proof-of-an-equation-involving-stirling-numbers-of-the-second-kind?noredirect=1&lq=1 or https://math.stackexchange.com/questions/1467247/an-identity-on-stirling-number-of-the-first-and-the-second-kind – delivery101 Dec 18 '19 at 08:28
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    Your attempt seems fine to me. – Angina Seng Dec 18 '19 at 08:36

1 Answers1

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If you are interested in an algebraic proof, it can be done as follows. Goal is

$$\sum_{m=k}^{n-l} {n\choose m} {m\brace k} {n-m\brace l} = {k+l\choose l} {n\brace k+l}.$$

The LHS is

$$\sum_{m=k}^{n-l} {n\choose m} m! [z^m] \frac{(\exp(z)-1)^k}{k!} (n-m)! [w^{n-m}] \frac{(\exp(w)-1)^l}{l!}.$$

This is

$$n! [w^n] \frac{(\exp(w)-1)^l}{l!} \sum_{m=k}^{n-l} w^m [z^m] \frac{(\exp(z)-1)^k}{k!}.$$

Now we may extend $m$ beyond $n-l$ to infinity because if $m\gt n-l$ the least power of $w$ that appears is $l+m \gt n$ (this is due to $\exp(z)-1 = z + \cdots$) which does not contribute to the coefficient extractor $[w^n]$ in front. We obtain

$$n! [w^n] \frac{(\exp(w)-1)^l}{l!} \sum_{m\ge k} w^m [z^m] \frac{(\exp(z)-1)^k}{k!}.$$

Note that $(\exp(z)-1)^k$ starts at $z^k,$ so the remaining sum copies the term in $z$ to a term in $w$ and we find

$$n! [w^n] \frac{(\exp(w)-1)^l}{l!} \frac{(\exp(w)-1)^k}{k!} \\ = n! [w^n] {k+l\choose l} \frac{(\exp(w)-1)^{k+l}}{(k+l)!} \\ = {k+l\choose l} {n\brace k+l}.$$

This concludes the proof. What we have used here is a so-called annihilated coefficient extractor (ACE).

Marko Riedel
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