Cube Digit Pairs

Each of the six faces on a cube has a different digit ($0$ to $9$) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of $2$-digit numbers.
For example, the square number $64$ could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: $01$, $04$, $09$, $16$, $25$, $36$, $49$, $64$, and $81$.
For example, one way this can be achieved is by placing $\{0, 5, 6, 7, 8, 9\}$ on one cube and $\{1, 2, 3, 4, 8, 9\}$ on the other cube.
However, for this problem we shall allow the $6$ or $9$ to be turned upside-down so that an arrangement like $\{0, 5, 6, 7, 8, 9\}$ and $\{1, 2, 3, 4, 6, 7\}$ allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain $09$.
In determining a distinct arrangement we are interested in the digits on each cube, not the order.
$\{1, 2, 3, 4, 5, 6\}$ is equivalent to $\{3, 6, 4, 1, 2, 5\}$
$\{1, 2, 3, 4, 5, 6\}$ is distinct from $\{1, 2, 3, 4, 5, 9\}$
But because we are allowing $6$ and $9$ to be reversed, the two distinct sets in the last example both represent the extended set $\{1, 2, 3, 4, 5, 6, 9\}$ for the purpose of forming $2$-digit numbers.
How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?

To form all the nine square numbers less than 100, you will need the digits 0, 1, 2, 3, 4, and then either 5 or 6, 8, and 9.

Let’s divide the problem into 2 cases:

Case 1: Both the cubes have the number 6 or 9:
From each cube, we choose 5 numbers from {0, 1, 2, 3, 4, 5, 8}. Two elements of 6 and 9 are already on both cubes so we don’t need to worry about them. There are 7 choices of 5 numbers or (7 choose 5 = 21) and we have two cubes, this gives a total of 21*21 = 441 arrangements. But, we have double-counted the cases that are similar, so we need to divide this number by 2 to correct the over counting, leading 441/2 = 220.5.

Case 2: Only one cube has the number 6 or 9:
From one cube, we choose 5 numbers from {0, 1, 2, 3, 4, 5, 8} and from the other one, we choose 5 numbers from {0, 1, 2, 3, 4, 8}. This gives a total of (7 choose 5)*(5 choose 5) = 21 arrangements.

The total number of distinct arrangements is the sum of the arrangements from the two cases:

220.5 (because we can’t have half an arrangement, it’s rounded up to 221) + 21 = 242 distinct arrangements.

It is important to note that there is a bit of number theory thinking involved in solving this problem and it is crucial to break it down into manageable cases and subtly use combinatorics to reach the solution.

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