Conceptually, what is $\langle x^k \rangle \big( \frac{1}{1-x} \big)^n$ with respect to $\binom{n+k-1}{k}$?

Question

If $\langle x^k \rangle f(x)$ denotes the k-th coefficient of the Taylor expansion around $0$, then

$$\langle x^k \rangle \Big( \frac{1}{1-x} \Big)^n=\binom{n+k-1}{k}$$

Note: Rewrite as $(1-x)^{-n}$ gives $\frac{1}{k!}(-1)^k\prod_{m=1}^{k}(-n-m+1$) and multiply in the $(-1)^k$

I know that the RHS can be interpreted as lining up n-1 uncolored and k colored balls such that no two colored ones are adjacent.

But what does the LHS mean conceptually/combinatorically?

Answer 1

Let $$\text{SEQ}(1)=\{\varepsilon,1,11,111,\ldots\}$$ denote all strings containing only $1$s with length $\geq0$. The empty string is denoted with $\varepsilon$. The corresponding generating function is $$x^0+x^1+x^2+\ldots=\frac{1}{1-x},$$ with the exponent of $x^n$ marking the length $n$ of a string of $1s$ and the coefficient of $x^n$ marking the number of strings of length $n$.

The generating function \begin{align*} \left(\frac{1}{1-x}\right)^n \end{align*}

corresponds to the $n$-fold cartesian product \begin{align*} \text{SEQ}^n(1) = \underbrace{\text{SEQ}(1)\times\text{SEQ}(1)\times\cdots\times\text{SEQ}(1)}_{n\text{ times}} \end{align*}

Identifying each element of the sequence $\text{SEQ}(1)$ with its length, the elements of $\text{SEQ}(1)$ become the non-negative integers and the elements of $\text{SEQ}^n(1)$ are $n$-tupels having non-negative integers as coordinates.

We conclude: The number of $n$-tuples having non-negative integers as coordinates which sum up to $k$ is given by \begin{align*} [x^k]\left(\frac{1}{1-x}\right)^n \tag{1} \end{align*}

We could also state (1) is the number of points in an $n$-dimensional hypercube having distance $k$ from a vertex when using the taxicab metric.