## Ring of Beads, Solution

Ok, so here’s the long awaited solution.

Let us suppose for a moment that in fact all the beads are of the same colour, i.e. indistinguishable. Then we know that they will return to the same configuration: any collision of two beads may as well be thought of as beads passing straight through each other so we will have the same configuration after one second.

Now to the problem. Here we *can* distinguish between beads. Nonetheless, by the previous argument, we know that after a second, the configuration of the beads will be the same apart from some of the beads perhaps being swapped. Well performing this permutation of beads again and again must lead to the original position at some point. Et voila.

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