To save myself some typing I am going to call the coins P, N, D, Q and L.
I am going to show you two ways to approach this problem.
Suppose I am going to use one coin to pay. That could be P, N, D, Q, or L. Thus there are five ways I could pay with one coin.
Suppose I am going to pay with four coins. I can do this by deciding which coin I am not going to use. There are also five choices for the coin not to include and hence there are five ways I can pay with four coins.
Suppose I am going to pay with two coins. If one of the coins you use is P then the coins you use could be PN, PD, PQ or PN for a total of 4 possibilities. If you don't use the P then there are four coins left and if you use the N coin then the coins you use could ne ND, NQ or NL for a total of 3 possibilities. Continuing this way I get 4 + 3 + 2 + 1 = 10 possibilities for two coin combinations.
Suppose I am going to use 3 coins then I could do this by deciding what 2 coins not to include. Thus again there are 10 possibilities for 3 coin combinations.
Suppose I am going to use all 5 coins. There is only one way to do this.
Thus I have found 5 + 5 + 10 + 10 + 1 = 31 different amounts.
Suppose you are going to look at an amount to pay. Consider the 5 coins, one at a time and decide if you are going to use it or not. Start with the P coin. You can use it or not so there are 2 choices. One you have made this decision look at the N coin. Regardless of what decision you mane with the P coin you have two choices for the N coin. Thus, so far there are $2 \times 2 = 2^2$ possibilities. Now the D. You can include it or not so for 3 coins you have $2 \times 2 \times 2 = 2^3 $ possibilities. Continuing this way I get, for all five coins, $2^5$ possibilities.
But $2^5 = 32$ not $31.$ Why the different answers in the two methods? Which is the correct answer?