Perhaps check it again

No, I would have no reason to consider the [placement of the] heads, because it is a matchsticks puzzle. The matchsticks puzzles have never limited themselves to placement of the heads. If you had wanted that extra condition, then you are changing what kind of traditional puzzle this is and not announcing in the rules.

Incorrect, TimeSpaceLightForce. In my explanation and diagram, I already showed that

Here is a diagram of the solution:

** This is my actual answer for the puzzle:

The ways that 6 moves that begin and end at point A can be diagrammed: where the moves happen to be completely to the left of A where the moves happen to be completely to the right of A where the moves start to the left of A but then eventually cross A before ending at A where the moves start to the right of A but then eventually cross A before ending at A Does anyone see more ways (paths) for any of the four categories above? I am not seeing any more than what I listed in the spoilers.

gavinksong, your number of ways for the six moves versus mine is off by two. I won't state whether it's by lower or higher. I don't see your method as fitting here. However, maybe it (your use of taking combinations) can be used as part of a larger different method. The problem is still open.

<------- g ------- f ------- e ------ d ------ c ------ b ------ A ------- B ------- C ------ D ------- E ------ F ------ G ------> A particle originates at point A and can move left one point or move right point for any move. It cannot be stationary for any move. For two moves, the ways can be diagrammed like this: ABC ABA Abc AbA Two of those ways begin and end with the particle at the starting point A. Question: For six moves, how many of the ways begin and end with the particle at the starting point A?

-1 for being an actual non-apology -2 for deflecting onto me -3 for continuing to be an actual non-apology -4 for continuing to deflect onto me - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

"Composite" refers to the number having at least two distinct positive integer divisors.

"I'm getting quite a collection of bonanova stars. (emoticon with big grin) Well, if you don't toot your own horn, others probably won't come along and do it for you. Someone should send you some ointment to put on your back where you have been frequently patting it.

I stated "No line can pass through all the interiors of the the three triangles." BMD, you stated "Why cant [sic] a line pass through [triangle] ABE and the common point of [triangles] ACD and EDF?" In my attempt to work on the problem, I misunderstood "crossing a polygon" from the original post and changed the subject. But, with you asking me that question (in the quotes just above), you changed the subject. I never addressed/denied that situation. It's not a line passing through the interiors of three triangles anyway.

I didn't question that "do not intersect each other" means that they do not overlap, meaning that the [line] segments do not have any points in common. But I can always have a situation where condition 2 occurs. That is why I claim there are other words/conditions/restrictions/qualifiers left out.

I claim it's not true, and I show it with a counterexample. In the xy-plane, let the following points be labelled as such: A = (0, 1) B = (1, 1) C = (-1, 0) D = (0, 0) E = (1, 0) F = (0, -1) Triangle ABE shares point A with triangle ACD. Triangle ACD shares point D with triangle EDF. Triangle ABE shares point E with triangle EDF. No line can pass through all the interiors of the three triangles.

BMAD, it looks like your problem statement is incomplete. I read it as "Given 50 [line] segments on the [a] line." It seems you left out a key word or words, because these line segments could be intersecting each other, or not intersecting each depending on where they are.