Stuck on generating operating functions.

Question

Find the ordinary generating function associated with the problem of finding the number of solutions in nonnegative integers of the equation

$$2a + 3b + 2c + d = r.$$

Answer 1

Given any finite sequence of positive integers

$$[k] \stackrel{def}{=} (k_1, k_2, \ldots k_m), \quad k_1, k_2, \ldots, k_m \in \mathbb{Z}_{+}$$

Let $\mathcal{N}_{[k]}(r)$ be the number of non-negative integer solutions for the equation

$$k_1 x_1 + k_2 x_2 + \ldots + k_m x_m = r, \quad x_1, x_2, \ldots, x_m \in \mathbb{N}$$

Let $f_{[k]}(t)$ be the corresponding ordinary generating function:

$$f_{[k]}(t) = \sum_{r=0}^\infty \mathcal{N}_{[k]}(r) t^r$$

If we have two finite sequences of positive integers $[k] = (k_1, k_2,\ldots,k_m)$ and $[\ell] = (\ell_1, \ell_2,\ell,\ell_m)$, we can concatenate them to form another finite sequence $$[k\ell]\stackrel{def}{=} (k_1,k_2,\ldots,k_m,\ell_1,\ell_2,\ldots,\ell_n)$$

The corresponding generating functions simply multiply. More precisely.

$$f_{[k\ell]}(t) = f_{[k]}(t) f_{[\ell]}(t)$$

When the finite sequence $[k] = (\alpha)$ consists of a single entry, i.e the underlying equation has the form $\;\alpha x = r\;$, the corresponding generating function is $$f_{(\alpha)}(t) = 1 + t^\alpha + t^{2\alpha} + \cdots = \frac{1}{1 - t^\alpha}$$
This implies the generating function for the equation $$\color{red}{2 a} + \color{orange}{3 b} + \color{green}{2c} + \color{blue}{d} = r$$ is simply $$ \left(\color{red}{\frac{1}{1-t^2}}\right) \left(\color{orange}{\frac{1}{1-t^3}}\right) \left(\color{green}{\frac{1}{1-t^2}}\right) \left(\color{blue}{\frac{1}{1-t}}\right) = \frac{1}{(1-t)(1-t^2)^2(1-t^3)}$$