But why is this a legitimate way of writing out trinomial expansion?

Question

So recently, I asked this question asking if $\sum_{r=0}^n\sum_{s=0}^r\dfrac{n!a^{n-r}b^{r-s}c^s}{s!(n-r)!(r-s)!}$ was a legitimate way of expanding $(a+b+c)^n$

Multi-binomial theorem

When working in more than two dimensions, it is often useful to deal with products of binomial expansions. By the binomial theorem this is equal to$$(x_1+y_1)^{n_1}\dots(x_d+y_d)^{n_d}=\sum_{k_1=0}^{n_1}\dots\sum_{k_d=0}^{n_d}\binom{n_1}{k_1}x_1^{k_1}y_1^{n_1-k_1}\dots\binom{n_d}{k_d}x_d^{k_d}y_d^{n_d-k_d}$$

Now say I wanted to write out $(a+b+c)^n$ in this way. Would I use this theorem to write it out as$$\sum_{r=0}^n\sum_{s=0}^r\dfrac{n!a^{n-r}b^{r-s}c^s}{s!(n-r)!(r-s)!}$$

Which then I explained how the expansion would be achieved. Now here's my question: Why is this a legitimate way of writing out the expansion of the trinomial theorem? I understand that yes, it is, and I understand the multinomial theorem, but would somebody be able to give me a combinatorial explanation of the formula?

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Why my previous question does not answer my question

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The reason that this is is because I had only received confirmation that yes, what I had reached as an expansion of the trinomial theorem was in fact equivalent to my original expression, but had not clarified why.

Answer 1

In the expansion of $(a+b+c)^n$, each term is of the form $ka^pb^qc^r$, where $k$ is the coefficient of $a^pb^qc^r$. In other words, $k$ is the number of ways to obtain $p$ number of $a$'s, $q$ number of $b$'s, and $r$ number of $c$'s from $(a+b+c)^n$. We can think of this process as follows: Write $(a+b+c)^n$ as $$ (a+b+c)_1(a+b+c)_2\cdots(a+b+c)_n $$ where we label each parenthesis a number. We now choose one of $a,b,c$ from each parentheis. Then $k$ will be the number of distinct ways to do this such that $p$ of $a$'s, $q$ of $b$'s, $r$ of $c$'s are chosen. This is because each term (without collecting like terms together) corresponds uniquely to one such way to choose.

The problem then becomes simple. We clearly have $$ k=\frac{n!}{p!q!r!}. $$ (Permutations of $n$ objects, modding out by permutations of the same terms chosen.) Summing over all indices $(p,q,r)$, we have $$ \begin{align*} (a+b+c)^n &= \sum_{p+q+r=n} \frac{n!}{p!q!r!}a^pb^qc^r \\ &= \sum_{i=0}^n \sum_{j=0}^i \frac{n!}{i!j!(n-i-j)!}a^ib^jc^{n-i-j} \\ &= \sum_{\ell=0}^n \sum_{j=0}^\ell \frac{n!}{j!(n-\ell)!(\ell-j)!}a^jb^{n-\ell}c^{\ell-j} \\ &= \cdots \end{align*} $$ There are many many different representations of the sum. It depends on how you loop through the $p,q,r$.