The question is as follows:
There are $n$ beads on a circular ring ($n \in \mathbb N$). At time $0$ they begin moving, some clockwise and some anticlockwise, but all with speed $1$. Whenever two beads collide, they collide elastically so that their speeds are unchanged and their directions are reversed. Does there exist a time $t$ at which all the beads are in exactly the same position as they were at time $0$?
I saw this question a while ago and attempted it - however, I'm not sure whether the question implies that the beads are distinguishable. If the beads aren't distinguishable, then the problem is fairly straightforward to solve...
Yes - whenever two beads are about to collide, imagine that they pass through each other instead. To an outside observer, since the beads are not distinguishable, it will look exactly the same as if the beads collided, and since each bead will be back at its original position after one revolution around the ring, such a time $t$ definitely exists.
... and so I was wondering if the problem can still be solved if the beads are distinguishable.
Thanks a lot for your help!
The answer is yes. Each time the indistinguishable case returns to its intial state, the distinguishable case is in a cyclic permutation of its initial state. It's cyclic because the beads don't pass through one another. After some number of cycles, you'll be back to the initial state itself.