So, the problem here is a combinatorics problem, finding the number of possible combinations with the following rules:
- There are 4 corners
- Each corner can have 0 to 6 dots
- Each of these dots can itself contain 0 to 6 subdots
- Each of these subdots can contain 0 to 6 sub-subdots
- The figure cannot be rotated
- The dots, subdots, or sub-subdots arrangements cannot be rotated
To get the answer:
First, we have to identify the number of different possibilities inside a dot. Which is 7 (one for each variant of a dice side, plus 0).
Then we identify the number of possibilities for the corners, which also gives us 7.
We then combine both results and work with combinatorics, in the case that there are no dots inside of the corner, there's only 1 ($7^0$) possible outcome. Only 7 if there is a single dot ($7^1$), 49 if there are two ($7^2$)...
This gives us the first part of our result, $\sum_{i=0}^6 7^{i}$. This result is the number of combinations for 0 to 6 dots, containing themselves 0 to 6 dots.
Then, after figuring out the sum for one corner, we have to apply it to all four corners.
If we do a small test and try with a single dot (only two possibilities, on and off), we have $2$ combination for a single corner, $4$ combinations for two corners, $8$ for three, and $16$ for all four corners.
We can easily see that the number of possible outcomes is equals to $2^n$ every time, $n$ being the number of corners. But in our case, there is not a single dot, so there are more than 2 possibilities. So we input our sum for how many possibilities a single corner has.
And we add the exponent: $(\sum_{i=0}^6 7^{i})^4 $
That gives us two possible "layers" of dots, but we're still missing another layer.
To add our deeper layer, we need to use our first formula $\sum_{i=0}^6 7^{i}$, which gives us the number of possible combinations for two layers, yet we need a third one. There can be up to 6 dots in one corner, having no dots will give us a single combination. Having only one dot gives us the sum we have found previously (let's call this sum $S$). We can, once more, identify a pattern, no dots is $S^0$, one dot is $S^1$ and so on...
This gives us $\sum_{j=0}^6 S^{j}$.
From there on, we just need to add our exponent for the four corners: $(\sum_{j=0}^6 S^{j})^4$
Which, when we replace $S$, gives us:
$$(\sum_{j=0}^6 (\sum_{i=0}^6 7^{i})^{j})^4$$
With a new result of: $$1999128512159148793087094162761709640097773563451792957120099762048442316537660592483207958099422011576770606002265990526401$$
(or just $1.99 * 10^{123}$ for a smaller result)
