Answer:

X^Y (X to the power of Y).

Where X is the number of different amino acids you will allow (2 or 10 in the specific question), and Y is the length of your polypeptide (dipeptide means only 2 long). You also need to define if you can re-use the amino acid, i.e. if the two amino acids can be the same. If we assume they can, then:

2^2 = 4

and

10^2 = 100

If you don’t allow the same amino acid to be re-used, you need to subtract every case where that is possible. For dipeptides, you need to subtract the number of different amino acids, X. In that case you can make 2 or 90 different dipeptides.

To make this more accessible, you can look at the example with 2 different amino acids, let’s call the “A” and “B”. If you have a dipeptide, then you have a “chain” that is 2 amino acids long. With 2 different to choose from, your possibilities are:

AA : AB

BA : BB

So 2^2 = 4. As you can see, AA and BB occur. As I said, you need to define whether this is OK or not (e.g. to you only have one of each amino acid, or do you have an infinite number, but only two different ones).

If you had three different ones, “A”, “B”, and “C”, then you could have:

AA : AB : AC

BA : BB : BC

CA : CB : CC

Which is 3^2 = 9. If on the other hand you had a tripeptide, with 3 different amino acids:

AAA : AAB : AAC :: ABA : ABB : ABC :: ACA : ACB : ACC

BAA : BAB : BAC :: BBA : BBB : BBC :: BCA : BCB : BCC

CAA : CAB : CAC :: CBA : CBB : CBC :: CCA : CCB : CCC

Making 3^3 = 27 combinations.

You can make your own table for the condition with 10 different amino acids to verify that this is actually true.