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 $k=0\dots 12$.