Enumerative Combinatorics problem from Stanley

Question

Let $n$ and $k$ be two positive integers. Show that \begin{align} \sum a_1\cdot a_2 \cdot \cdots \cdot a_k={n+k-1\choose 2k-1} , \end{align} where the sum ranges over all compositions $(a_1,a_2,\ldots,a_k)$ of $n$ into $k$ parts.

Reminder: A composition of $n$ into $k$ parts is defined as a $k$-tuple $(a_1,a_2,\ldots,a_k)$ of positive integers such that $a_1 + a_2 + \cdots + a_k = n$.

This is Exercise 1.29 in Stanley's Enumerative Combinatorics volume 1 (2nd edition). I have no idea how to do it.

Answer 1

It suffices to prove they have the same generating function:

\begin{align*} \sum\limits_{n} \sum a_1\cdots a_k x^n &= \sum\limits_{1\leq a_1,\ldots,a_k} a_1\cdots a_k x^{a_1 + \ldots + a_k} \\ &= \sum\limits_{1\leq a_1,\ldots,a_k} (a_1 x^{a_1}) \cdots (a_k x^{a_k}) \\ &= \left(\sum\limits_{1\leq a_1} a_1 x^{a_1}\right)\cdots\left(\sum\limits_{1\leq a_k} a_k x^{a_k}\right) \\ &= \left( \frac{x}{(1 - x)^2} \right)^k \\ &= x^k \left( \frac{1}{(1 - x)^{2k}}\right) \\ &= x^k \sum\limits_m \binom{m + 2k - 1}{m}x^m \\ &= x^k \sum\limits_n \binom{n - k + 2k - 1}{n - k}x^{n-k} \\ &= \sum\limits_{n} \binom{n + k - 1}{2k - 1}x^n. \end{align*}

Answer 2

You have a row of $n$ green olives on the table. You insert $k-1$ toothpicks into the row, dividing it into $k$ non-empty segments. Each of the $\binom{n-1}{k-1}$ ways of doing this corresponds to one of the $\binom{n-1}{k-1}$ compositions of $n$ into $k$ parts: if $a_i$ is the number of olives in the $i$-th segment, the composition is $a_1+\ldots+a_k=n$.

Now consider the olive-and-toothpick row corresponding to the composition $a_1+\ldots+a_k=n$; there are $a_i$ ways to choose an olive from the $i$-th segment and replace it with a black olive, so there are $a_1a_2\ldots a_k$ ways to replace one olive from each segment with a black olive. Thus, the lefthand side of your identity is the number of ways to insert the $k-1$ toothpicks and then replace one olive from each of the resulting $k$ segments with a black olive.

Alternatively, we could start with a string of $n+k-1$ positions on the table and choose $2k-1$ of them, say positions $p_1<p_2<\ldots<p_{2k-1}$. We’ll put a toothpick in each of the $k-1$ positions $p_2,p_4,\ldots,p_{2k-2}$ and a black olive in each of the positions $p_1,p_3,\ldots,p_{2k-1}$, and we’ll put a green olive in each of the remaining $(n+k-1)-(2k-1)=n-k$ positions. We then have one of the possible outcomes of the first procedure, and every outcome of the first procedure can be obtained uniquely in this way, so we’ve simply counted the same strings of olives and toothpicks in a different way. Clearly this procedure can be carried out in $\binom{n+k-1}{2k-1}$ ways, so the identity is established.

Added: A proof by induction on $k$ is also possible. Let $\varphi(n,k)$ be the sum on the lefthand side of the identity. Then $\varphi(k,k)=1$ for all $k$, and for any $n>k$ we also have the recurrence

$$\varphi(n,k+1)=\sum_{\ell=1}^{n-k}\ell\varphi(n-\ell,k)\;.$$

Suppose that

$$\varphi(n,k)=\binom{n+k-1}{2k-1}$$

for all $n\ge k$, and let $n\ge k+1$. Then

$$\begin{align*} \varphi(n,k+1)&=\sum_{\ell=1}^{n-k}\ell\varphi(n-\ell,k)\\ &=\sum_{\ell=1}^{n-k}\ell\binom{n-\ell+k-1}{2k-1}\\ &=\sum_{\ell=2k-1}^{n+k-2}(n+k-1-\ell)\binom{\ell}{2k-1}\\ &=(n+k)\sum_{\ell=2k-1}^{n+k-2}\binom{\ell}{2k-1}-\sum_{\ell=2k-1}^{n+k-2}(\ell+1)\binom{\ell}{2k-1}\\ &=(n+k)\binom{n+k-1}{2k}-2k\sum_{\ell=2k-1}^{n+k-2}\binom{\ell+1}{2k}\\ &=(n+k)\binom{n+k-1}{2k}-2k\binom{n+k}{2k+1}\\ &=(2k+1)\binom{n+k}{2k+1}-2k\binom{n+k}{2k+1}\\ &=\binom{n+k}{2k+1}\\ &=\binom{n+(k+1)-1}{2(k+1)-1}\;, \end{align*}$$

and the result follows by induction.