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Order does matter so this is a permutation problem. Since we have no repeating elements, then use the standard formula for selecting K elements from a pool of N elements:

Yes, since you are only picking numbers, picking ABC is supposedly to be the same with picking CBA, so it is a combination problem. We have C(5,3)=5!/2!3!. For easier calculation I always do this: C(5,3)=C(5,2)=5*4/2!=10

If the question asks how many different three letter words can you form, then order matters since ABC is a different word from CBA. In that case the answer would be P(5,3)=5*4*3=60. _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

It does not matter in what order letters are chosen, so it's a combination problem. if each letter can be chosen only once, then C(5,3)=5!/3!(5-3)!=10 _________________