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Q&A

How can a 15 year old construe the LHS of Generalized Vandermonde's Identity, when it lacks summation limits and a summation index?

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Paradoxically, though Rothe-Hagen Identity (henceforth RHI)

$\sum\limits_{k=0}^n\frac{x}{x+kz}{x+kz \choose k}\frac{y}{y+(n-k)z}{y+(n-k)z \choose n-k}=\frac{x+y}{x+y+nz}{x+y+nz \choose n}$

generalizes Generalized Vandermonde's Identity (henceforth GVI),

$\sum\limits_{k_1+\cdots +k_p = m} {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p} = { n_1+\dots +n_p \choose m }$

RHI is more intelligible than GVI for my 15 year old. A 15 y.o. can effortlessly write any term of RHI, by substituting the lower limits for all $k$ in the addend. When $k = 0$, just input $k = 0$ in the addend. When $k = n$, just swap all $k$'s in the addend with $n$'s!

But how can a 15 y.o. interpret the LHS of GVI? Or even write the first few terms of the LHS of GVI? It contains no lower and upper limits of summation, and no summation index. GVI contains no $k$, unlike RHI!

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Rothe-Hagen is a distraction (1 comment)

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Unfortunately, this is ambiguous notation. It isn't your fault; this presentation of this identity is more suited for an audience already familiar with the material than someone learning it for the first time.

The answer to your question is that the summation is over all ordered length-$p$ sequences of non-negative integers such that the sequence sums to $m$. Inside the summation, a single one of these sequences is represented with the variables $k_1, k_2, \ldots k_p$.

The detail that the summation is over ordered sequences and not unordered partitions is important, but there is nothing in the notation itself that requires this interpretation. One way a reader of that Wikipedia article [the OP originally linked here in their question, and still uses the presentation from that source] could understand that ordered sequences are meant is by constructing the analogy with the original Vandermonde's identity, which I repeat below (with some convenient renaming of variables):

$$ \sum_{k = 0}^m {n_1 \choose k}{n_2 \choose {m - k}} = {{n_1 + n_2}\choose m} $$

This is the $p = 2$ case of the generalized identity. Notice, for example, that the cases being summed include both a $k = 0$ term (which simplifies to ${n_1 \choose 0}{n_2 \choose m}$) and a $k = m$ term (which simplifies to ${n_1 \choose m}{n_2 \choose 0}$). These two terms in the generalized identity would correspond to two length-2 sequences both summing to $m$: $(0, m)$ and $(m, 0)$. Since both of these terms are included, we can infer that the order of the sequences in the generalized identity must be important.

Different mathematicians use different notations for things like sequences, but if I were to take a stab at a more precise notation for generalized Vandermonde, it would be something like the following:

$$ \sum_{(k_i) \in \mathcal{K}_m^p} {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p} = { n_1 + \ldots + n_p\choose m} $$

and separately define $\mathcal{K}_m^p$ to be the set $\{(k_i)\mid(k_i)\in\mathbb{N}^p \wedge \sum_{i=1}^p k_i = m\}$. ($\mathbb{N}^p$ is common notation for the set of all length-$p$ sequences, or $p$-tuples, of elements of $\mathbb{N}$; and $(k_i)$ is common notation for talking about an entire sequence as opposed to the $i$th element of the sequence.)

From this version of the notation, it's clear that $m$, $p$, and of course the $n_i$ are fixed, and that the $k_i$ vary. It's also explicit in the definition of $\mathcal{K}_m^p$ that we are summing over ordered sequences of non-negative integers of length $p$.

Wikipedia can be a great resource for learning math, but in this regard I'm afraid it's let you down. Better luck next time!

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Thanks. How can you rewrite the Generalized Vandermonde's Identity with a Summation Index, and Upper ... (2 comments)
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You can rewrite it in different notation with sum limits if you want. You just need to use a different way to express the constraint on which terms to include in the sum. E.g. with the Iverson bracket notation the LHS becomes

$$\sum_{k_1 = 0}^m \sum_{k_2 = 0}^m \cdots \sum_{k_p = 0}^m [k_1+\cdots +k_p = m] {n_1\choose k_1} {n_2\choose k_2} \cdots {n_p\choose k_p}$$

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