In how many ways can 25 forints be paid using 1, 2, 5, 10, and 20 forint coins?
We can answer the question by counting the possible cases.

First, let's examine the case when we use only one type of coin for the payment. This obviously cannot be an even value coin (2, 10, 20 Ft), since 25 is odd. Thus, we have $\underline{2}$ possibilities, which are paying entirely with 1 Ft coins or entirely with 5 Ft coins.

Next, let's count the cases where we use two different denominations. This can be 1 and $2 \mathrm{Ft}$ coins in the following ways:

$1 \mathrm{db} \quad 1 \mathrm{Ft}$ and $12 \mathrm{db} \quad 2 \mathrm{Ft}$,

$3 \mathrm{db} \quad 1 \mathrm{Ft}$ and $11 \mathrm{db} \quad 2 \mathrm{Ft}$,

and so on, always decreasing the number of 2 Ft coins by 1 and increasing the number of 1 Ft coins by 2. This results in $\underline{12}$ cases.

When using 1 and 5 Ft coins, we start with 5 pieces of 1 Ft and 4 pieces of 5 Ft, and similarly, by decreasing the number of 5 Ft coins and increasing the number of 1 Ft coins, we have $\underline{4}$ possible cases.

It is easy to see that with 1 and $10 \mathrm{Ft}$ coins, the number of possibilities is 2, while with 1 and $20 \mathrm{Ft}$ coins and 5 and $20 \mathrm{Ft}$ coins, only $\underline{1}-\underline{1}$ case is possible. With 2 and $5 \mathrm{Ft}$ coins, as well as 5 and $10 \mathrm{Ft}$ coins, there are $\underline{2}-\underline{2}$ cases.

So far, this totals $2+12+4+2+1+1+2+2=26$ possibilities.

There is still the possibility of using 3 or 4 different denominations for the payment. Similarly to the previous cases, we can determine that in the first case, the total number of possibilities is $\underline{36}$, while in the second case, it is $\underline{6}$.

In the end, the total number of ways to pay $25 \mathrm{Ft}$ using 1, 2, 5, 10, and 20 Ft coins, assuming we have enough of each, is: $26+36+6=68$.