General form for this problem I encountered this problem by Polya about counting the number of ways that a dollar can be changed. Suppose that there are pennies (worth 1), nickels (worth 5), dimes (worth 10), quarters (worth 25) and fiftycent coins (worth 50). The number of ways to change a dollar (worth 100) can be written as the following generating function:
$$
D(z) = \sum_p z^p \sum_n z^{5n} \sum_d z^{10d} \sum_q z^{25q} \sum_f z^{50f}
$$
where $D(z)$ is:
$$
\frac{1}{(1-z)(1-z^5)(1-z^{10})(1-z^{25})(1-z^{50})}
$$
I understand the generating function, but is there a general form to express its coefficients given any set of denominations ? i.e. How to dervive $[z^n]D(z)$, where:
$$
D(z) = a_0z^0 + a_1z^1 + a_2z^2 + ...
$$
and the coefficient $a_k$ of $z^k$ express the number of ways that an amount worth $k$ can be arrived at given denominations $\{d_1,d_2,d_3,...,d_n\}$, i.e. in the example above, we have $n=5$ and $d_1=1,d_2=5,d_3=10,d_4=25,d_5=50$.
EDIT:
It looks like that a general form for this problem is hard to compute ... (the problem hints that a computer simulation may be needed) ... However, it looks like that $D(z)$ is asymptotic to the following formula, where $N$ represents the denomination of the bill, i.e. if it is a dollar, we have $N=100$
$$
\frac{N^{t-1}}{d_1d_2,...,d_t(t-1)!}
$$
Is there an explanation why $D(z)$ has this asymptotic form ?

If you have a closed formula for your generating function $D(z)$, all you have to do to obtained its coefficients differentiate several times and evaluate at zero.
This works since if
$$ D(z) = a_0 + a_1 z + a_2 z^2 + ...$$
then we have that
$$a_k = \frac{D^{(k)}(0)}{k!},$$
and this is easy enough to compute for finite $k$ with a bit of Maple/Mathematica/Sympy.
If this is not the kind of closed formula you're after, you will have to look at the comment of  Matti P and Somos to find a different closed expresion in terms of finite sums.
