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In a two-toss experiment, if (H, T) is considered different from (T, H), you are counting permutations. If they are considered the same, you are counting combinations. For permutations, there are 4 distinct outcomes: (H, H), (H, T), (T, H), (T, T). For combinations, there are 3 distinct...

Then you MUST study what Cantor did. You will find it beautifully logical. The set of Natural Numbers are countably infinite and any set that can be put into 1-1 correspondence with them is also countably infinite. The set of Rational Numbers is countably infinite. Any countably infinite union...

IMO, this discussion should focus on the OP, which specified the counting size of sets. Otherwise, it can become very unfocused.

The problem statement specifies that all failures are independent. All links on that path are up. It indicates nothing about the ADB path. That path could be up or down. Correct. This is irrelevant. No, The problem states that all probabilities are independent. No. No. You are making this...

One take-away from this is how rapidly a simple problem can become unmanageable. Combinatorial problems grow exponentially and the slightest modification can make the calculations very difficult. By "unmanageable", I mean it might take a supercomputer decades to solve.

I see. The solutions that depend on counting combinations of 3 lines don't work. That and the need for symmetry is key to most of the ideas here. The solutions are not very "robust". I have a general preference for robust approaches, but they are often not elegant.

It certainly complicates symmetry arguments. Maybe I misunderstand your first statement. I don't see that it invalidates anything about a triads of intersecting lines. Triangles still have 3 distinct sides.

On the subject of set sizes, it would benefit you to read about what Cantor did and how he approached the subject. IMHO, it has been very well thought out and is interesting.

I'm not sure what you mean here. 1) If an intersection is beyond the figure area, should it count as a possible triangle vertex or not? 2) Which solution are you calling the "alternative solution"?

IMHO, the desire for a non-"brute force counting" method is misguided here. The logic is so specialized and convoluted that it required several erroneous tries to create a "Rube Goldberg" logic that might work. The "brute force" method still requires some tricks, but the result is much more...

I wrote a program to do brute force counting and its answer matches that of @bob012345 in post #15. I have not thoroughly examined the results, but I have some realistic hope that my program is correct. Here is the program and results. Each triangle is identified by three vertices numbered as...

That seems logical. This is a treacherous problem. I would feel more confident of your logic if there was a simple, brute-force, counting that gave the same answer, but everything I think of quickly gets complicated.

I think that even brute force counting would be challenging. And being confident that a mathematical method works would be very hard unless the correct answer is known (probably from some trusted brute force counting.)

I would consider that to be automating a "brute force counting" method. IMO, although it is not what the OP asked for, it is still an essential step to verify the answer from an analytical method.

It's true of any music genre that time lets us remember the good ones and forget about the tons of mediocre/bad ones. Our memories are kind that way. ( I remember looking for deep meaning in "Rockin' Robin, and finding it!) You can find great new C&W songs if you look in the right places, for...