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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.

N. F. Taussig
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mahen3d
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  • Are you familiar with generating functions? – N. F. Taussig Jul 18 '22 at 02:00
  • nope, but some insight into how to solve the problem in general would be handy too, this is a year 3 kid question i knw – mahen3d Jul 18 '22 at 02:07
  • https://math.stackexchange.com/questions/2869279/find-no-of-non-negative-integer-solutions-of-a2b3c-200?rq=1 might help –  Jul 18 '22 at 02:11
  • This book has a nice recursive algorithm that solves that. You basically take the 2 dollar coin plus all the ways to make 3 dollars with all the coins plus the number of ways to make 5 dollars without the 2 dollar coin. In your case you must also take the extra step of eliminating solutions that violate your constraint on the number of each coin. – John Douma Jul 18 '22 at 02:31

2 Answers2

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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.

Cheese Cake
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    I might be wrong but the OP seems to try to know about a method other than counting all the possibilites. –  Jul 18 '22 at 02:17
  • I guess this is kind of "organised" guessing, but unfortunately I do not know any formulas. kodlu's answer is pretty good. – Cheese Cake Jul 18 '22 at 02:22
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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

Parcly Taxel
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kodlu
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