Thank you so much! Melody, your logic is correct!
This is what AOPS says :
Solution:
The quick way to tackle this problem is correcting for overcounting. There are 6!/6 = 5! ways to arrange 6 beads in a circle. Since we can flip the bracelet, we have 5!/2 = 60 arrangements. Dividing by 2 once more to account for the fact that 2 beads are the same, we have 60/2 = 30 final arrangements.
We can also break the problem into three cases: the identical beads are adjacent, there's one bead between the identical beads, and there's two beads between the identical ones (i.e. they're opposite each other).
If the identical beads are adjacent, there are 4!/2=12 ways to order the other beads, where we divide by 2 because the bracelet can be flipped over to give another one of the initial 4! orderings (so these two orderings are indistinguishable)---hence, our 4! counts each possible arrangement twice. Make sure you see why we aren't dividing by anything for rotation---we don't count rotations of the same arrangement multiple times in our initial count of 4!.
If the two beads are one apart, there are 4 ways to choose the bead that is between them, and ways to order the other three. As before, we can flip the bracelet to give another one of the initial orderings, so there are 4!/2=12 ways to order the beads this way.
Finally, we have the two beads opposite each other. As before, we have 4! initial orderings. However, this time, there are 2 symmetries---the bracelet can be flipped and it can be rotated 180 degrees (thus swapping the positions of the two identical beads). Hence, each distinguishable arrangement is counted 4 times, so there are 4!/4=6 orderings.
Hence, there are 12+12+6=30 distinguishable ways for her to put the beads on her bracelet. If you still don't quite follow, break out a keychain and six keys and try it yourself.
