Solution. For the change of 5000 Ft, the banknotes in circulation that can be used are 2000, 1000, 500, and $200 \mathrm{Ft}$ denominations. Let's denote the number of these in the sum by $a, b, c$, and $d$ respectively. We know that $a+b+c+d=10$, and

$$
2000 a+1000 b+500 c+200 d=5000
$$

Let's simplify by 100:

$$
20 a+10 b+5 c+2 d=50
$$

From this, it is clear that $d$ must be divisible by 5, so the value of $d$ can be $0, 5$, or 10.

If $d=10$, then $a, b$, and $c$ must be zero. Since $10 \cdot 200=2000$ is not a solution, $d$ must be either 0 or 5.

Let $d=0$ and divide equation (1) by 5:

$$
4 a+2 b+c=(a+b+c)+3 a+b=10
$$

Since $a+b+c=10$, from this we get $3 a+b=0$; with $a \geq 0, b \geq 0$, only $a=b=0$ is possible. Then from (2), $c=10$. This is a solution because $10 \cdot 500=5000$.

If $d=5$, then from (1) we get $20 a+10 b+5 c=40$. Simplify by 5, and replace $a+b+c$ with 5:

$$
4 a+2 b+c=(a+b+c)+3 a+b=5+3 a+b=8
$$

From this, $3 a+b=3$, so either $a=1$ and $b=0$, or $a=0$ and $b=3$. If $a=1, b=0$, and $d=5$, then $c=4$. The obtained decomposition is:

$$
1 \cdot 2000+4 \cdot 500+5 \cdot 200=5000
$$

If $a=0, b=3$, and $d=5$, then $c=2$ and

$$
3 \cdot 1000+2 \cdot 500+5 \cdot 200=5000
$$

With the available banknotes, we can pay 5000 Ft in a total of three different ways.
10 banknotes have a total value of $5000 \mathrm{Ft}$. What banknotes could make up this amount?