How many ways I can make \$5 in 50c, \$1 and \$2 coins? With five 50c coins, five \$1 coins, and five \$2 coins.
In how many different way can I make up $5? I need a equation or method to solve this problem rather than guessing with permutations.
 A: Basically, what I did was to start with two \$$2$ coins, and decrease the amount of \$$2$ coins.
\begin{array}{|c|c|c|}
\hline
$2& $1 & 50c & \text{Okay?} \\ \hline
2 & 1& -& yes\\ \hline
2 & - & 2& yes\\ \hline
1 & 3 & -& yes\\ \hline
1 & 2 & 2& yes\\ \hline
1 & 1 & 4& yes\\ \hline
1 & - & 6& no\\ \hline
- & 5 & -& yes\\ \hline
- & 4 & 2& yes\\ \hline
- & 3 & 4& yes\\ \hline
- & 2 & 6& no\\ \hline
- & 1 & 8& no\\ \hline
- & - & 10& no\\ \hline
\end{array}
I counted 8 ways, because some had more than 5 of the same type of coins, which is why there is an "okay" column in the table.
A: Working in cents, find the coefficient of $x^{500}$ in the product
$$
(1+x^{50}+x^{2\times 50}+x^{3\times 50}+x^{4\times 50}+x^{5\times 50})\times \\
(1+x^{100}+x^{2\times 100}+x^{3\times 100}+x^{4\times 100}+x^{5\times 100})\times \\
(1+x^{200}+x^{2\times 200}+x^{3\times 200}+x^{4\times 200}+x^{5\times 200})
$$
and since all your coins are divisible by 50 in denomination, you can divide by 50 in the exponents and look for the coefficient of $x^{10}$.
You can generalize to different quantities (say $m$ two dollar coins) of coins via
$$(1+x^{200}+x^{2\times 200}+\cdots+x^{m\times 200})$$
or unlimited supply of coins via an infinite series where you can usually ignore convergence properties
