I did a couple of problems on circular permutation, where clockwise and counterclockwise were distinguishable.
My textbook says if they aren't distinguishable, then it's $(n-1)!/2$ ... Well this makes sense. But then the problem went like, if 10 "different" beads are to be used to make a necklace, then the number of necklaces that can me made are (10-1)!/2
Wait now hold on! If all of those beads are different, I think this should be enough of a reason to think the CW and AW can be distinguished. So shouldn't it be simply (10-1)!= 9! ..?
I could understand if beads were all alike.
If my textbook's right, then could you explain me why? Also, what would you do, if the condition was all beads are alike?
Your textbook is right because you can turn the necklace over giving the same necklace, but with the beads essentially reversed. If all the beads are indistinguishable, well there's easily only 1 permutation.