How can I think of Multinomial Theorem's "urn" model by permutation language.

Question

Can you explain me a little how to think of the relation with (1) and the permutaion language

$$\begin{align*}\left(\begin{array}{c} n\\ k_1,k_2,\text{...},k_n \end{array}\right)=\frac{n!}{k_1!k_2!\text{...}k_m!}\tag{1}\end{align*}$$

statement1:

In permutaion language, (1) is the permutation number of $n$ elements that partitioned into $k$ tuples, in each tuples they are the same, but different with each other in different tuples. The number of elements in each tuples is $,n_1,n_2,\text{...},n_k$($n_1+n_2+\text{...}+n_k=n$).

In the special case, Binomial Theorem

$(a+b)^n=(a+b)(a+b)\text{...}(a+b)$, I know it's choosing k's $a$ from n tuples $(a+b)$(think it is a box, but according to the statement1, there are $a$ and $b$ in one box, so the box is not the same as that in statement1(the tuples))

$\left(\begin{array}{c} n\\ k \end{array}\right)a^kb^{n-k}$

Answer 1

My comment was badly worded. Here's the correct version.

In permutation language, $\binom{n}{k_1,k_2,\ldots,k_\ell}$ is the number of ways of arranging $n$ letters, with $k_1$ copies of letter $1,$ $k_2$ copies of letter $2$, $k_3$ copies of letter $3,$ and so on. For example, ABABC is one of the $\binom{5}{2,2,1}=\frac{5!}{2!\,2!\,1!}=30$ ways to arrange two As, two Bs, and one C.

There are two ways to think about this:

1. Arrange $n$ letters in $n!$ ways. Then, since permutations of repeated letters amongst themselves do not change the arrangement, divide by the number of such permutations.
2. An arrangement can be formed by regarding the letters as being placed into $n$ slots. Choose which $k_1$ of the $n$ slots are to contain letter $1$; then choose which $k_2$ of the $n-k_1$ remaining slots are to contain letter $2$; then choose which $k_3$ of the $n-k_1-k_2$ remaining slots are to contain letter $3$; and so on. There are $\binom{n}{k_1}\binom{n-k_1}{k_2}\binom{n-k_1-k_2}{k_3}\ldots$ ways to do this.

If you write out the binomial coefficients appearing in Method 2, you will find that the product telescopes (i.e. there is a systematic cancelation between one factor and the next), leaving the same result as Method 1.

Addendum in response to query in the comments:
A telescoping sum is one where there is cancelation between one term and the next, leading to a big simplification. They come up often in the study of sequences. An example would be $$\begin{aligned}\sum_{i=1}^5\frac{1}{i(i+1)}&=\sum_{i=1}^5\left[\frac{1}{i}-\frac{1}{i+1}\right]\\ &=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)\\ &=1/1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4)+(-1/5+1/5)-1/6\\ &=1/1-1/6.\end{aligned} $$ One can also have telescoping products. Let's actually try writing this one out: $$ \begin{aligned}&\binom{n}{k_1}\binom{n-k_1}{k_2}\binom{n-k_1-k_2}{k_3}\ldots\binom{n-k_1-\ldots-k_{\ell-1}}{k_\ell}\\ &\quad=\frac{n!}{k_1!\,(n-k_1)!}\frac{(n-k_1)!}{k_2!\,(n-k_1-k_2)!}\frac{(n-k_1-k_2)!}{k_3!\,(n-k_1-k_2-k_3)!}\ldots\frac{(n-k_1-\ldots-k_{\ell-1})!}{k_\ell!\,(n-k_1-\ldots-k_\ell)!}\\ &\quad=\frac{1}{k_1!\,k_2!\,\ldots\,k_\ell!}\cdot\frac{n!}{(n-k_1)!}\frac{(n-k_1)!}{(n-k_1-k_2)!}\frac{(n-k_1-k_2)!}{(n-k_1-k_2-k_3)!}\ldots\frac{(n-k_1-\ldots-k_{\ell-1})!}{(n-k_1-\ldots-k_\ell)!}\\ &\quad=\frac{1}{k_1!\,k_2!\,\ldots\,k_\ell!}\cdot\frac{n!}{(n-k_1-\ldots-k_\ell)!}\\ &\quad=\frac{1}{k_1!\,k_2!\,\ldots\,k_\ell!}\cdot\frac{n!}{0!}.\end{aligned} $$