How many combinations possible to make $10 from a set amount of coins

Question

I have 50 coins. They add up $10. They are comprised of 10 cent, 20 cent and 50 cent coins. How many combinations are possible?

I tried making equations and etc.

x + 2y + 5z = 100 cents
x + y + z = 50
y + 4z = 50 cents
x = 3z

However I don’t know what to do next. I’m expecting a quartic or some sort of polynomial, because there should be multiple combinations possible. However I don’t know how to go about finding it.

Answer 1

INTUITIVE APPROACH

First you can notice that $50$ coins of $20$c give you $10$\$. So you have at least one possibility. The other possibilities can be obtained from this one thinking that you have to change some of these $20$c coins in $10$c and $50$c coins. To be clearer you are allowed to change $n$ coins of $20$c, with $p$ coins of $10$c and $q$ coins of $50$c such that $p+q=n$ (the total number of coins is fixed) and the value of the $n$ coins of $20$c must be equal to the one obtained with the new coins. The minimum amount of coins $n$ that you can change is $4$. Indeed $4$ coins of $20$c are equivalent to $3$ coins of $10$c and $1$ coin of $50$p. Actually all the possible change you can make must be multiple of $n=4$. So you have $n=4$, $n=8$, $n=12$ and so on, up to $n=48$ (since $n\leq 50$). This means that you have $12$ possibilities. Adding the initial solution of only $20$c coins you get that there are $13$ possible combinations.

RIGOROUS APPROACH

Let $x$ be the number of $10$c coins, $y$ of $20$c and $z$ of $50$c. You are looking for the non-negative integer solutions of the system $$10x+20y+50z=1000\,;$$ $$x+y+z=50\,.$$

The solutions are in the form $$x=\frac{3}{4}(50-y)\,;\qquad z=\frac{1}{4}(50-y)\,.$$ So you are looking for the number of integer $y$ between $0$ and $50$ such that $50-y = 0\mod 4$. So how many $y\in\{0\dots 50\}$ such that $y\mod 4 = 2$? There are $13$ such $y$, precisely given by $$y = 2+4k$$ with $k=0\dots 12$.

Answer 2

$y+4z=50$ means that $4z\le 50$, so there are $13$ possible values for $z$. That's all you need, because given $z$, we are forced to choose $x=3z$ and $y=50-4z$. It is easy to check then that these values satisfy your first two equations.