How many distinct elements does a group of permutation on 3 letters have?

Question

I am having some problems solving a problem similar to this. So i tried making it a more simpler problem. I really don't know how to approach this kind of problem. A hint would be very much appreciated.

Answer 1

If I interpreted your question correctly, you want to know how many elements are in the group of all permutations of a 3-element set. The answer is $3! = 6$. This is a special case of a general result that says the group of all permutation of a set with $n$ elements has $n!$ elements.

The group of all permutations of a finite set with $n$ elements is known as the symmetric group, denoted $S_n$. The group you're asking about is the symmetric group $S_3$.

Edit: I just realized that you only asked for a hint. Sorry about that. To verify that the above answer is correct, try to actually write out all permutations of a 3-element set such as $\{a, b, c\}$ (by writing out all possible arrangements of those three letters). You should end up with six permutations.