There is a famous, well-known problem in the world of “rich math tasks” that involves taking an n x n x n cube and painting the outside of it. Then you break apart the large cube into unit cubes (see image below cribbed from here for n=2 and n=3):
Notice that some of the unit cubes have 3 painted faces, some have 2 painted faces, some have 1 painted face, and some have 0 painted faces.
The standard question is: For an n x n x n cube, how many of the unit cubes have 3 painted faces, 2 painted faces, 1 painted face, and 0 painted faces.
[In case you aren’t sure what I mean, for a 3 x 3 x 3 cube, there are 8 unit cubes with 3 painted faces, 12 unit cubes with 2 painted faces, 6 unit cubes with 1 painted face, and 1 unit cube with 0 painted faces.]
Earlier this year, I worked with a middle school student on this question. It was great fun, and so many insights were had. This problem comes highly recommended!
Today we had some in house professional development, and a colleague/teacher shared the problem with us, but he presented an insight I had never seen before that was lovely and mindblowing.
Spoiler alert: I’m about to give some of the fun away. So only jump below / keep reading if you’re okay with some some spoilers.
First off, here are some manipulatives we used in our PD:
We used colors to represent the number of painted faces. So pink = three painted faces, blue = two painted faces, yellow = one painted face, green = zero painted faces.
And we, though standard arguments, saw that for an n x n x n cube
green cubes (zero painted faces) =
yellow cubes (one painted face) =
blue cubes (two painted faces) =
pink cubes (three painted faces) =
There are so many ways to come up with those individual formulas.
And clearly these add up to the total number of cubes. So we have:
Lovely. (Feel free to expand each term out to see that the left side and right side truly are equal.)
Here’s the insight my colleague shared with us.
If you look carefully at the right hand side, you can actually see as the same thing as .
By applying the binomial expansion to , each term in the expansion gives us the number of cubes which have 3 sides painted, 2 sides painted, 1 side painted, or 0 sides painted. Seriously, use the binomial expansion on and see for yourself! WHAAAAA?
So here’s the question he posed to us, and I pose to you…
Why? Why the heck does each term in the binomial expansion for give you the number of cubes with 3 sides, 2 sides, 1 side, and 0 sides painted?
I mean, we know it works algebraically. We can expand it out to see. But why, conceptually? Where do the and fit into everything?
When I finally figured it out, my mind was blown. So simple and elegant, yet so unintuitive for me. I’m not going to type out my insight here. Yet, anyway. Because I want to leave it for others to think through. But if there is interest, after a short while, I can definitely do an update with the conceptual insight!