In how many ways can I pay the bill with different type of money?

Question

I how many ways can I buy fish for 10 euros if I have?: 5€ - 3pcs 2€ - 6pcs 1€ - 2pcs

I try with primitive aproach:

5+5=10------1

5+2+2+1-----2

2+2+2+2-----3

2+2+2+2+1+1-4

Is there any other combination? How to calculate that?

Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. You must be able to figure out some ways to pay the bill, for example. — saulspatz, Mar 17 '20 at 17:11
The problem is small enough that you should be able to brute force it. Break into cases on the number of $5$'s used, then on the number of $1$'s used, noting that there must be an even number of odd numbers used to have the total be $10$. So, you have $5+5, 5+1+2+2$ and $2+2+2+2+1+1$ and $2+2+2+2+2$ and you are done. If you are wanting a "more mathematical approach" then look at the expansion of $(1+x^5+x^{10})(1+x^2+x^4+x^6+\dots+x^{12})(1+x+x^2)$ and the coefficient of $x^{10}$ there. — JMoravitz, Mar 17 '20 at 17:17

Mathsmerizing · Answer 1 · 2020-03-17T17:38:21.553

0

We need to find the coefficient of $x^{10}$ in the $(1+x^5+x^{10}+x^{15})(1+x^2+x^4+...+x^{12})(1+x+x^2)$

coefficient of $x^{10}$ in the $(1+x^5+x^{10}+..)(1+x+x^2)(1+x^2+x^4+...+x^{12})$

coefficient of $x^{10}$ in the $(1+x+x^2+x^5+x^6+x^7+x^{10}+..)(1+x^2+x^4+...+x^{12})$

$=1.x^{10}+x^2.x^8+x^6.x^4+x^{10}.1=4$

edited Mar 17 '20 at 17:38

answered Mar 17 '20 at 17:18

Mathsmerizing

Your first line is correct. Everything that follows is suspect. There is no way the result is $30$. – JMoravitz Mar 17 '20 at 17:22
Got it. I didn't think it through. I took $\frac{1}{1-x}$ for every expression – Mathsmerizing Mar 17 '20 at 17:28
Thanks for that! The proof that for every problem there is always more then one solution. Can this be solved with combinatorics? – Cigls Mar 17 '20 at 19:24

score 0 · Accepted Answer · answered Mar 17 '20 at 17:59

Another method: Number of non-negative integral solutions of the equation $$5x+2y+z=10; x \le 3, y\le 6, z\le2$$

Case i: $x=0 \Rightarrow 2y+z=10 \Rightarrow z\in$ even so either z=0, y=5 or z=2, y=4

Case ii: $x=1 \Rightarrow 2y+z=5 \Rightarrow z\in$ odd so z=1, y=2$

Case iii: $x=2 \Rightarrow 2y+z=0 \Rightarrow z=0, y=0$

Four solutions $(0,5,0),(0,4,2),(1,2,1),(2,0,0)$

2 Answers2