On a cuboidal planet with side length 1, either people live on each of the eight vertices. One day, each of them decides to move to a vertex that is a distance of $\sqrt{2}$ away from him. What is the probability that no two of them occupy the same vertex after this move?
People on a Cube
6
$\begingroup$
combinatorics
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0@MarkDominus: that's right, it would b$e$ one per vertex. But eight is closer to either. – 2012-07-19
1 Answers
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Hint: you can color the vertices with two colors. People stay on their own color and can move to any other vertex of that color. So you can consider the problem on a tetrahedron, and square the result. Then you are looking for a derangement on the four items (why?)