Starting with three different amino acid mole | vedclass

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Question

(D) Let the three different amino acids be $A$,$B$,and $C$.For a tripeptide formed by three different amino acids,the number of possible arrangements

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Let the three different amino acids be $A$,$B$,and $C$.For a tripeptide formed by three different amino acids,the number of possible arrangements is given by the permutation of $3$ distinct items taken $3$ at a time.The number of arrangements is $3! = 3 	imes 2 	imes 1 = 6$.The possible tripeptides are: $ABC$,$ACB$,$BAC$,$BCA$,$CAB$,and $CBA$.Thus,$6$ different tripeptide molecules are formed.

Starting with three different amino acid molecules,how many different tripeptide molecules are formed?

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