The trick to this problem is realising that by turning a "6" upside down, we can make a "9". Since there are only a maximum of 31 days in a month, we will never need a 6 and a 9 simultaneously.
Because each month as an 11th and a 22nd, both the cubes will need to have the numbers 1 and 2 on them. Also, because each of the digits needs to appear next to a zero (for the first 9 days of the month), zero should also be on both cubes.
This leaves us with three free spots on each cube, where we need to put the numbers 3, 4, 5, 6, 7, and 8.
So, C_{1} : 0, 1, 2, 3, 4, 5
C_{2} : 0, 1, 2, 6, 7, 8 |