counting

5 choices for first position

   4 choices remain for second position

       3 choices remain for third position                         5  x  4  x 3  = _________   ways

I suppose the order doesn't matter, so I'd expect the answer to rather be 5C3 (5 choose 3 and not 5 permute 3), which gives 5!/(2!*3!) = 10 ways.