I am able to come up with good generating functions to solve combinatorial problems, but unable to appreciate its usefulness. For example, I have following combinatorial problem.
How many ways to pay $50$ using unlimited $1, 2 , 5$ coins, (1) when order matters, (2) when order does not matter.
My generating functions:
(1) Number of ways = coefficient of $x^{50}$ in $(1+x+x^2+x^3+...+x^{50})(1+x^2+x^4+x^6+...+x^{50})(1+x^5+x^{10}+x^{15}+...+x^{50})$
Till now, it's fun. But now what? Calculating the coefficient of $x^{50}$ from above expression is as good as counting all individual combination of $1,2,5$ to make $50$, manually. So how is generating function useful in counting?
For case (2) I have following generating function.
Number of ways = coefficient of $x^{50}$ in $(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...+\frac{x^{50}}{50!})(1+x^{2}+\frac{x^{4}}{2!}+\frac{x^{6}}{3!}+...+\frac{x^{50}}{25!})(1+x^{5}+\frac{x^{10}}{2!}+...+\frac{x^{50}}{10!})$
However this permutes only same coins. Unable to come up how all types of coins will be permuted among themselves. Also, same question like above, how generating function is helping in count? I can very well count it manually, like $5$ can occur in $0$ to $10$ times, and for each of these occurrences of $5$, count occurrences of $2$ and $1$.
I would say your generating function in (1) is actually the answer to the "order does not matter" question. A simple function which also works is $\dfrac{1}{(1-x)(1-x^2)(1-x^5)}$. This gives a sequence (starting at zero money) of $1,1,2,2,3,4,5,6,7,8,10,\ldots$ ways of paying coins.
For order mattering the numbers should be higher, you can use the generating function $\dfrac{1}{(1-x-x^2-x^5)}$. This gives a sequence (starting at zero money) of $1,1,2,3,5,9,15,26,44,75,128,\ldots$ ways of paying coins.
In both cases you can turn these into a recurrence for calculation, respectively
each starting $a_0=1$ and $a_n=0$ for $n<0$.