$$ 1 + x + x^2 + \dots + x^n = \frac{x^{n+1}-1}{x-1}$$
$$ x^{a} + x^{a+1} + \dots + x^b = x^a(1 + x + x^2 + \dots + x^{b-a}) = \frac{x^a(x^{b-a+1} -1)}{x - 1} = \frac{x^{b+1} - x^a}{x-1}$$
The range of integers $a_k$ to $b_k$ are represented in the powers of $x$ of the generating function.
Multiply all the generating functions for the ranges $x_1$ to $x_m$ together.
When multiplied the powers of each term are added to the power of other terms in different ranges.
$$F(x) = \prod_{n=1}^{m} \left( \frac{x^{b_n + 1} - x^{a_n}}{x-1} \right) = \prod_{n=1}^{m} \left( x^{a_n} + x^{a_n + 1} \dots + x^{b_n} \right) = \sum_{k}c_k x^k$$
$c_n$ is the number of ways that the sum (power of $x^n$) can add up to $n$.
i.e. the powers add up to the sum $n$ and the coefficient $c_n$ is the count of how many ways that sum can be made.
Try a very small example by hand e.g. a pair of dice.
Each dice has a range of six numbers $1$ to $6$ so how many ways can $7$ be made.
$$F(x) = (x^1 + x^2 + \dots + x^6)(x^1 + x^2 + \dots + x^6) $$
The left () is one dice the right () is the other. Multiply them to find the combinations.
$$ F(x)= x^{12}+2\,x^{11}+3\,x^{10}+4\,x^9+5\,x^8+6\,x^7+5\,x^6+4\,x^5+3\,x^
4+2\,x^3+x^2$$
There are $6$ ways to roll $7$ : $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$.
Note there is no way to roll $1$ with a pair of dice.