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We have a set of coins with values 1, 2, 5, and 10, and banknotes with values 5 and 10. We want to pay a bill of 15 using a combination of these coins and banknotes. How many ways can we pay the bill?

I attempted to formulate the problem as follows:

$$x_1+2x_2+5x_3+10x_4=15,\quad x_i\ge0$$

and then proceeded to count the solutions manually. However, this method is time-consuming Can you provide any suggestions or tips on how to approach this problem more efficiently

Ariana
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    I've read that but didn't understand that. And I don't understand how this problem related to polynomials. – Ariana Feb 18 '24 at 07:18
  • Just to be clear: You describe a system of 6 different items. Du you have an unlimited number of each of these items.? – DJohnM Feb 18 '24 at 07:33
  • yes there are unlimited – Ariana Feb 18 '24 at 08:36
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    @Ariana -- "I don't understand how this problem related to polynomials." Then it's surprising you attempted to approach the problem with a polynom. Which, BTW, is incomplete, the banknotes are missing me think. – m-stgt Feb 18 '24 at 08:53
  • @m-stgt but the similar questions tried to solve it with polynomials. I think we can count banknotes later for example if we choose a coin with value 10 we must count it twice because we can choose 10 valued banknote instead of coin – Ariana Feb 18 '24 at 10:11
  • This problem differs from the usual change making problem by introducing both coins and bills for the $$5$ and $$10$ denominations. I'm inclined to allow the Question in order for that small variation to be treated by a new Answer. – hardmath Feb 19 '24 at 16:58
  • @Ariana Set up two different but identical polynomials for the two ways to get ten and five, respectively. The polynomials will be the same but there will be two copies of each in the product you use to compute the final coefficient of $x^{15}$. – John Douma Feb 20 '24 at 14:05

1 Answers1

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Let $f_2(n)$ be the number of ways to get $n$ by using just the $1$ and $2$ denominations. I'll leave figuring it out to you.

If we only had one choice for the $5$ denomination, we would choose how many we want to use, and then use $f_2$ to count the number of ways to make up the remainder:

$$f_5(n) = \sum_{0 \le 5k \le n}f_2(n-5k)$$ But in our problem with two types of $5$s, if we use $k\ 5$s, we have $k+1$ choices of how many of the $k\ 5$s are coins ($0$ up to $k$), so $$\bar f_5(n) = \sum_{0 \le 5k \le n}(k+1)f_2(n-5k)$$

With $10$s a similar thing happens, so we have $$\bar f_{10}(n) = \sum_{0 \le 10k \le n}(k+1)\bar f_5(n-10k)$$

In particular, $$\bar f_{10}(15) = (0+1) \bar f_5(15) + (1 + 1)\bar f_5(5)$$

And I'll leave the rest to you.

Paul Sinclair
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  • @Ariana -Though you accepted this as the right answer, as I review it now, I notice a mistake. If the $5$ denominations used were all distinguishable, then it would be correct that for $k$ of them, there would be $2^k$ choices. But since they are not distinguishable, the only choice is how many of the $k$ should be coins, and there are $k+1$ choices ($0$ to $k$). I will correct it. – Paul Sinclair Feb 22 '24 at 12:35