Christmas calendar

You've probably seen one of these.  This one was on display at my mother-in-law's house this year.  I've never really looked closely at one to figure out how they work, and I decided not to look at this one either.  Rather, I determined the distribution of the numbers on the two cubes while driving around my hometown during the holidays.  My parents live at one end; my in-laws live at the other.  So, lots of opportunity.

The first observation to make is that because of the 11th and 22nd, both cubes must have a 1 and a 2.  If you don't think deeply at all beyond that you leap to an incorrect conclusion: since there are 8 faces remaining and you have 8 numbers left to distribute (0 and 3-9) you can just put half of them on one cube and half on the other.  But that's a problem.  If 0 is on only one face then you can get only 6 of the single digit dates (the 6 numbers on the cube without a 0).  It follows that you must put a 0 on both cubes along with the 1 and the 2. 

But wait, that leaves 7 digits to place on the remaining 6 faces.  If we weren't using Arabic numerals we'd be in trouble, but luckily the numbers we use have the (in this case) useful typographical property that the 9 is an upside-down 6 (or vice versa, if you prefer).  And, since we don't need the 6 and the 9 at the same time for this application, we can safely put only the digits 3 through 8 on the remaining 6 faces.  Voila! Calendar cubes.

I realize this is utterly trivial, but it's about as much brain power as I can muster while on vacation.  A more interesting question is: how many such cubes can there be?  That leads to all kinds of interesting combinatorics.  Note, though, that determining which three digits you place on the first cube determines those on the second cube.  Choose three digits from the set of six possibilities; there are \(\binom{6}{3} = 20 \) ways to do this.  We then have a set of six numbers: \( \{0,1,2,a,b,c\} \).  How many different cubes can we make labeled with these digits?  This is a classical combinatorics question, usually phrased in terms of paint colors.  There are \( 6! \) ways to place the digits on the cube, but we need to consider rotational symmetry: there are 24 rotations of a cube and so there are 720/24 = 30 such cubes.  So, for each selection of three digits we get 30 distinct first cubes and 30 distinct second cubes, making 900 pairs of cubes for each of the 20, giving us a grand total of 18,000 pairs of calendar cubes.  But wait, I counted each pair twice, so there are only 9,000.  Still quite a few, but I'm sure they only come in one variety.

Maybe next time I'll just look at the cubes.