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I need help proving this. Let $n$ and $k$ be positive integers, and let $$\left(\!\!{n\choose k}\!\!\right) ={n+k -1\choose k},$$ prove:

$$\left(\!\!{n\choose k}\!\!\right) = \sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right).$$

I'm thinking I need to show that the equation $$\sum_{i=0}^{k}\ \left(\!\!{n-1\choose i}\!\!\right) = {n+k -1\choose k}$$ is true but I am unsure how to make this jump.

Anne Bauval
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H. Madland
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3 Answers3

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$$ \begin{align} \sum_{i=0}^k\left(\!\!\binom{n-1}{i}\!\!\right) &=\sum_{i=0}^k\binom{n+i-2}{i}\tag{1a}\\ &=\sum_{i=0}^k(-1)^i\binom{1-n}{i}\tag{1b}\\ &=(-1)^k\sum_{i=0}^\infty\binom{-1}{k-i}\binom{1-n}{i}\tag{1c}\\ &=(-1)^k\binom{-n}{k}\tag{1d}\\[3pt] &=\binom{n+k-1}{k}\tag{1e}\\[3pt] &=\left(\!\!\binom{n}{k}\!\!\right)\tag{1f} \end{align} $$ Explanation:
$\text{(1a):}$ definition
$\text{(1b):}$ negative binomial coefficient
$\text{(1c):}$ $(-1)^k\binom{-1}{k-i}=(-1)^k(-1)^{k-i}[k\ge i]=(-1)^i[i\le k]$
$\text{(1d):}$ Vandermonde's Identity
$\text{(1e):}$ negative binomial coefficient
$\text{(1f):}$ definition

robjohn
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This is related to the Hockey-stick identity.

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I encourage the reader to find out how...

Here's a combinatorial proof.

  1. Let’s say there is a $2$-D plane where we have demarcated $k+n$ number of points.
  2. We want to form as many clusters of $n+1$ points as possible.
  3. Clearly, there’s going to be $\displaystyle \binom{k+n}{n+1}$ clusters.
  4. Now, we need to dive into the world of every individual point.
    1. The first point has to choose $n$ points from remaining $k+n-1$ , that makes $\displaystyle \binom{k+n-1}{n}$ clusters
    2. The second point will have to choose $n$ points from remaining $k+n-2$ points. (Because we are not dealing with the first one again.) That makes $\displaystyle \binom{k+n-2}{n}$ clusters.
    3. And this goes on till the last point has exactly $n$ points remaining to choose from. Hence$\displaystyle \binom{n}{n}$ cluster(s).
  5. Thus in total we have these many ways: $\displaystyle \binom{k+n-1}{n} +\binom{k+n-2}{n} +\binom{k+n-3}{n} +\ldots +\binom{n}{n}\tag*{}$ Flipping the series in (5) and equating with (3): $\displaystyle \binom{n}{n} +\binom{n+1}{n} +\binom{n+2}{n} +\ldots +\binom{n+k-1}{n} =\binom{n+k}{n+1}\tag*{} $ $\displaystyle \text{or, }\sum _{i=0}^{k-1}\binom{n+i}{n} =\binom{n+k}{n+1}\tag*{} $ $\displaystyle \text{i.e., }\boxed{\sum _{i=0}^{k-1}\binom{n+i}{i} =\binom{n+k}{k-1}}\tag*{} $

Take $k=k'+1$ and $n=n'-2$. $\blacksquare$

Nothing special
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Combinatorial Proof: Suppose we have n objects in set $S = \{1, 2, \cdots, n\}$.

LHS: $\left(\!\!\binom{n}{k}\!\!\right)$ is the number of $k$-subsets from $S$ with repeats (by definition of multichoose).

RHS: We consider the number of occurrences of $n$ in a k-subset of $S$.

  • If $n$ is not in the subset, we have $\left(\!\!\binom{n-1}{k}\!\!\right)$ such subsets since we create a $k$-subset from all objects except $n$.

  • If $n$ is in the subset once, we have $\left(\!\!\binom{n-1}{k-1}\!\!\right)$ such subsets since we have $k-1$ spots remaining in the subset.

  • If $n$ is in the subset twice, we have $\left(\!\!\binom{n-1}{k-2}\!\!\right)$ such subsets.

    $\vdots$

  • If $n$ is in the subset $n$ times, we have $\left(\!\!\binom{n-1}{0}\!\!\right) = 1$ such subsets since no spots remain.

Thus, $\sum_{i=0}^k \left(\!\!\binom{n-1}{i}\!\!\right)$ represents all such cases when $n$ is in the k-subset $0, 1, 2, \cdots, n$ times.

Bryce
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