# How many different tiles are there when each corner may have 0-6 dots, each of which may have 0-6 dots?

There are four corners to each tile. Each corner can be empty, or contain an arrangement of dots (1-6) like the sides of a dice. Within each of these dots can be a further arrangement of 1-6 dots, or no dots. Then once again, within those dots, there can be a further arrangement of 0 to 6 dots (see picture above). The tiles have a unique orientation, they cannot be rotated. Same for the dots, they cannot be rotated (there is only 1 orientation possible for all different numbers of dots).

How many different tile arrangements are there?

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

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)

• What about symmetries? Commented Apr 12, 2022 at 19:58
• I don't think symmetries are important, but if they are (and OP mentions so) I can change it Commented Apr 12, 2022 at 21:01
• How can they not be? If I can turn one tile to look like another are they a different arrangement? What if all the possible tiles are in a bag, and I pull two identical tiles out, how are you going to tell the difference between the two arrangements you pulled out? Commented Apr 12, 2022 at 23:05
• Small clarification for OP. This is a result and not a function, it does take into account the possibility of no corners at all, or any number of the four corners being absent. Commented Apr 13, 2022 at 13:02
• clever clever. upvote for you, m8
– user79541
Commented Apr 19, 2022 at 14:48

Total Unique tiles: 39858898360862731501938422350790258474401

It this worth writing up or deleting @Auribouros?

• Your answer assumes that every subsequent set of dots is the same as the parent one. For a single dot in a single corner, it will go like this: You can have 0 dots $(7^0)$, 1 dots, containing from 0 to 7 dots $(7^1)$, 2 dots, containing from 0 to 7 dots $(7^2)$... and so on until six dots, and that's only for one dot in one corner Commented May 2, 2022 at 13:31
• So would multiplying the totals for each set of dots, rather than adding, solve that? Commented May 2, 2022 at 13:49
• It won't, you need the sum of all possible combinations, which also happens to be a sum, feel free to check my answer for the logic used. Commented May 2, 2022 at 13:51
• Don't forge to mark an answer as correct if it fits what you seek/answered your question Commented May 2, 2022 at 15:17
• I forge in order to gain knowledege :) nothing else. Commented May 3, 2022 at 10:41