Regarding usage of series for computing number of integrals solutions, in combinatorics.

Question

I was recently studying a theorem called Multinomial Theorem, in combinatorics, the theory regarding the topic was given in my book as follows,

Consider the equation $$x + y+ z = 4$$ and that we have to find the number of non-negative integrals solution to this equation, every single variable can take values from $0$ to $4$, both included, let their values be set as the exponent of some quantity say, $p$, for each variable the set of terms with different exponents of $p$ is $\{p^0, p^1, p^2, p^3,p^4, p^4\}$

The number of all possible triplets is the no. of terms in the product $$(p^0 + p^1 + p^2 + p^3 + p^4 + p^4)^3$$ Each term for a variable i.e. $x,y,z$.

To find the number of integral solutions for the equation we must find the coefficient of $p^4$ in the expansion of the above polynomial, hence we have to figure out the coefficient of $p^4$ in $$(1-p^5)^3(1-p)^{-3}$$

or its coefficient in $(1-p)^{-3}$.

Till this point in the theory I understood everything quite well, now the textbook put forward the taylor series expansion of $(1-x)^{-n}$ to help us solve the problem,

$$(1-x)^{-n} = \sum_{k = 0}^{\infty}{n +k-1\choose k}x^k$$

Now I have studied calculus, so taylor series are not new to me, however now the textbook proceeded to find the coefficient of $p^4$ using this taylor expansion, and stating that as the answer.

What i do not understand is that why does this work?, how do the coefficients of the taylor series actually answer the problem, nowhere in computing the taylor series was i adding up exponents of the various terms then why does this even work.

Any help would be highly appreciated!.

Answer 1

It is hard in math to answer the question "Why does this work?". The best I can do is to show you the rigorous proof for why this works, and hope that showing the details gives you some intuition.

Lemma: Let $k\in \mathbb N$. Let $S_1,\dots,S_k$ be subsets of the nonnegative integers. Let $a_n$ be the number of solutions to the equation $$x_1+\dots+x_k=n, \text{ such that}\\x_i\in S_i\text{ for }i\in \{1,\dots,k\}.$$Finally, define $f_i$ to be the analytic function defined by $f_i(x)=\sum_{k\in S_i}x^k$. Then $$ \sum_{n=0}^\infty a_nx^n=f_1(x)\times f_2(x)\times \dots \times f_k(x). $$

First, I will prove this in the case where $k=2$, because the notation gets cumbersome as $k$ increases. $$ f_1(x)\times f_2(x)=\left(\sum_{k\in S_1}x^k\right)\left(\sum_{\ell\in S_2}x^\ell\right)=\sum_{k\in S_1}\sum_{\ell\in S_2}x^{k+\ell} $$ Now, let us find the coefficient of $x^n$ in the double sum $\sum_{k\in S_1}\sum_{\ell\in S_2}x^{k+\ell}$. Note that there is a contribution of $+1$ to the coefficient of $x^n$ for each ordered pair $(k,\ell)$ such that $k\in S_1$ and $\ell\in S_2$ where $k+\ell=n$. But these ordered pairs exactly correspond to solution to the equation $x_1+x_2=n$ with $x_1\in S_1$ and $x_2\in S_2$. We conclude that the coefficient of $x^n$ is equal to $a_n$ as defined in the statement of the Lemma.

For general $k$, the proof idea is the same. Instead, you have a $k$-fold product of generating functions, which turns into $k$-iterated sum. $$ f_1(x)\times \dots \times f_k(x)=\sum_{\ell_1\in S_1}\sum_{\ell_2\in S_2}\dots\sum_{\ell_k\in S_k} x^{\ell_1+\ell_2+\dots+\ell_k} $$ Again, there is a contribution to $x^n$ for each ordered $k$-tuple $(\ell_1,\dots,\ell_k)$ such that $\ell_1+\dots+\ell_k=n$ with $\ell_i\in S_i$, which corresponds exactly to the number of solutions to the equation $x_1+\dots+x_k=n$.