Today I was looking for another fun problem and found another problem that I thought would make a fun project:

Barbara has an octahedron, and she wants to color its vertices with two different colors. How many different colorings are possible? By “different” we mean that you can’t make one look like the other throu a re-orientation.

I started by introducing the problem and asking the kids what their initial ideas were:

They had a couple of pretty good ideas including some basic ideas about symmetry. Using those ideas we began counting the different colorings:

We counted the cases in which 3 vertices were black and 3 vertices were red. This case proved to be tricky, but going through it slowly got us to the correct answer.

Finally, as a fun little extension, I asked them to find the number of ways to color the faces of a cube with two colors. Having solved the octahedron problem already, this one went pretty quickly, and they even noticed the connection between the two problems 🙂

I like this problem. I’m glad that the boys were able to see some of the basic ideas. When you add more colors the counting gets much more difficult and some pretty advanced math comes into play. The number of colorings with “n” colors is:

The different terms correspond to different symmetries of the cube / octahedron. We’ll have to wait a few more years to cover the complete details 🙂