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I know this question has been asked before and has been answered here and here.

I have a slightly different formulation of the Hockey Stick Identity and would like some help with a combinatorial argument to prove it. First I have this statement to prove: $$ \sum_{i=0}^r\binom{n+i-1}{i}=\binom{n+r}{r}. $$ I already have an algebraic solution here using the Pascal Identity: $$ \begin{align*} \binom{n+r}{r}&=\binom{n+r-1}{r}+\binom{n+r-1}{r-1}\\ &=\binom{n+r-1}{r}+\left[\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-1)-1}{r-2}\right]\\ &=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\left[\binom{n+(r-2)-1}{r-2}+\binom{n+(r-2)-1}{(r-2)-1}\right]\\ &\,\,\,\vdots\\ &=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-2)-1}{(r-2)-1}+\binom{n+(r-3)-1}{r-3}+\cdots+\left[\binom{n+1-1}{1}+\binom{n+1-1}{0}\right]\\ &=\binom{n+r-1}{r}+\binom{n+(r-1)-1}{(r-1)}+\binom{n+(r-2)-1}{(r-2)-1}+\binom{n+(r-3)-1}{r-3}+\cdots+\binom{n+1-1}{1}+\binom{n-1}{0}\\ &=\sum_{i=0}^r\binom{n+i-1}{i}. \end{align*} $$

I have read both combinatorial proofs in the referenced answers above, but I cannot figure out how to alter the combinatorial arguments to suit my formulation of the Hockey Stick Identity. Basically, this formulation gives the "other" hockey stick. Any ideas out there?

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Laars Helenius
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2 Answers2

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Note that $\binom{n+r}{r}=\binom{n+r}{n}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$. On the other hand, for $i=0,1,2,\ldots,r$, $\binom{n+i-1}{i}=\binom{n+i-1}{n-1}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$ whose largest element is $n+i$.

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Problem Statement

For non-negative integers $n$ and $r$, prove that:

$$ \sum_{i=0}^r \binom{n+i-1}{i} = \binom{n+r}{r} \quad \text{(Expression I)} $$

Proof Strategy

I claim that each sides of Expression I, counts the number of ways to distribute up to $r$ indistinguishable objects into $n$ distinguishable bins. Therefore, the sides are equal and the identity holds.

This can also be seen as the number of non-negative integer solutions to the Diophantine inequality: $$ x_1 + x_2 + \ldots + x_n \leq r $$ where integers $x_1, x_2, \ldots, x_n$ represent the number of objects in each of $n$ bins.

Let us define $A(n, r)$ as the number of non-negative integer solutions to the Diophantine inequality.

I will compute $A(n, r)$ using two different ways and show that each of them yield one side of the Expression I, thereby proving the desired identity by double counting.

Counting In Two Ways

Method 1: Converting The Inequality to An Equality

Introducing an auxiliary variable $x_{n+1}$, we rewrite the Diophantine inequality as an equation:

$$ x_1 + x_2 + \ldots + x_n + x_{n+1} = r $$

where the non-negative integer $x_{n+1}$ represents the number of "slack" objects, i.e. number of objects that are not distributed in the bins. As the solution to the stars and bars problem goes, the number of solutions is:

$$ A(n, r) = \binom{n+r}{r} \quad \text{(Expression II)} $$

Method 2: Considering All Cases

Since all desired answers of the Diophantine inequality are integers, let us consider all possible exact sums from $0$ to $r$:

$$ x_1 + x_2 + \ldots + x_n = i \quad \text{for each } i \in \{0, 1, \ldots, r\} $$

where the $i$-th equation corresponds to $i$ objects being distributed ($i \in \{0, 1, \ldots, r\}$).

Now the inequality is converted to several Diophantine equations. The number of solutions to the $i$-th equation, counted similar to the stars and bars problem, is:

$$ \binom{n+i-1}{i} $$

By the principle of addition, summing over all these configurations, we have:

$$ A(n, r) = \sum_{i=0}^r \binom{n+i-1}{i} \quad \text{(Expression III)} $$

Conclusion

By equating the results of the two methods, i.e. Expression II and Expression III, we conclude:

$$ \binom{n+r}{r} = \sum_{i=0}^r \binom{n+i-1}{i} $$

This completes the proof by double counting.

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