There is a misconception that runs through the questions, the answer key and the comments: For tripeptides, the peptide linkage is directional (i.e. the tripeptide has an amino terminal and a carboxy terminal), so ABC is different from CBA.
To get the number of tripeptides combined from three types of amino acids, you have three choices for the first, three independent choices for the second and another three independent choices for the third position, for a total of 3 x 3 x 3 = 27 possibilities.
[OP] I am unsure if I've done it in a valid way, as answer given is 18.
The answer key is incorrect. I have no way of reading the question that 18 would be the answer, even considering the poor wording of the question.
[OP] For AAZ type, we've three set, AAB, ABA, AAC, ACA. So, total 12 tripeptides.
There are three choices for Z, two remaining choices for A, and three positions for Z (AAZ, AZA, ZAA). So, total of 3 x 2 x 3 = 18 tripeptides containing two different amino acid types of the initial set of three types.
[Orthocresol] This question is phrased very poorly. Without reading your working, I would have answered 20×19×18=6840.
This answers how many different tripeptides made from (20) standard amino acids contain no amino acid twice. The question, certainly poorly stated, says "synthesized by using three different amino acids". So your reagents are a set of three amino acids types, and you are making tripeptides from them.
[Maurice] With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.
That is indeed a math problem, not a biochemistry problem. You can make AAA from A and B, but also from A and C, so you counted AAA twice (and BBB and CCC). So your answer overshot by 3.
[Rahul Verma] Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides?
The linkage does not introduce new chiral centers (different from e.g. combining sugars), so each sequence XYZ designates a single molecule (given that you start with enatiomerically pure amino acids).
[porphyrin] The list is 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 which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.
The list is correct, but there is no reason to remove palindromes (9 in the list) or divide non-palindromes by two (18 in the list).
I missed all the comments that came in while I was writing this. They now reflect the misconceptions and how these can be resolved; what is not resolved is the poorly worded question and its incorrect answer key - maybe someone who has the original question can take on the task of sending an erratum.