Thanks, @apm. :)
I’ve gone over all of your drawings and have been trying to determine if it’s a complete set. I think you nailed it. There may be small A intersecting medium B and both inside big C that’s not included, but I could be completely wrong here.
I think this is #75 in the lower left corner?
I’ll admit that I am a bit lost with your math below. Can you help me out? I’m a tad rusty on combinatorics, to be honest, but interested to understand.
Sure. If we have three circles that interact pairwise, that makes six different interactions (AB, AC, BA, BC, CA, CB) - this is often modeled as drawing two of three balls from an urn without putting them back, with three options for the first ball and two for the second (3*2=6).
In this case, we're not interested in the order, though, so we need to divide that by the number of possible permutations of each pair: 6/2=3 (basically, we want to treat AB as being the same as BA). This means that, between three circles, there are three "interactions" (or, as we called them earlier, "2-circle configurations").
Now we want to find out what combinations of interactions are possible. There are five different types of interactions (1.separate, 2.touching outside, 3.intersecting, 4.touching inside, 5.contained). Here, we want to allow duplicates, so we can enumerate 125 possible combinations (5*5*5=125). Again, we're not interested in the order of interactions (e.g. we don't want to deal with CBA if we've already dealt with ABC), so we need to trim down that list a bit.
Here, it helps to look at three different types of combinations.
1. All interactions are different. This is similar to the above example of picking three of five balls from an urn (5*4*3), then dividing by the number of permutations of each result (3*2*1). This leads to ten different results.
2. Two interactions are the same, the third one is different. This is like picking two of five balls (5*4) and then magically duplicating the first ball. Because we're doing this, we mustn't divide by the number of permutations (2*1) in this case - instead, we're considering all twenty different results.
3. Last but not least, the results where all three interactions are the same. With five different interactions possible, the result here is obviously five as well.
Altogether, that's 10+20+5 = 35 potential results, of which one seems to be impossible.
For my images, I did exactly this. Enumerate all 125 possibilities
1. Separate - Separate - Separate
2. Separate - Separate - Touching Outside
3. Separate - Separate - Intersecting
4. Separate - Separate - Touching Inside
5. Separate - Separate - Contained
125. Contained - Contained - Contained
Then trim down to unique combinations, then try to draw an image for each. That's where the strange numbering in the upper left corner of each square is coming from.