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.

harey, if you'll just finish the problem, you've got it. The problem does not have to be solved by a system of equations, but because your post involved that, I am posting continuing work under this spoiler for you. Here is a system of simultaneous equations:

Regarding post #2, you have a certain amount of doubt regarding correctness and/or justification of certain steps, as evidenced by question marks (as a for instance). I am looking for a solution by anyone made with complete confidence, and where the steps are justified from the prior ones. I'm not wanting any steps that come about through luck/inspection/trial and error.

Can the following be completely factored over integer coefficients, where the highest degrees on x and y are both one? 2x2 + 3xy - 2y2 - x + 3y - 1 If not, state that. If so, then show the steps leading up to and the final factored form.

witzar, I clicked yours as "Mark Solved," because it's posted before the next poster's, and you both show the same formula.

To follow up to my prior post, I will show the listing and the total for n = 5:

bonanova, the main puzzle asks for the answer for when n = 4. Maybe you kept it to yourself. Did you have the value for n = 4? Pickett, you didn't get a confirmation either way for the total for n = 4, before going into more calculations for the totals for larger n and then working on a formula. The values for n = 4, 5, ... are different. In the following spoiler, I show the listing and the total for n = 4:

Let n = a positive integer. Suppose you choose n distinct points on a line. The distance between any pair of points is irrelevant. A "curve" is either: 1) an individual point 2) a set of at least two points 3) an individual line segment 4) a set of at least two line segments 5) some combination of the above I will label the points with A, B, C, ..., Y, Z, AA, AB, ..., and so on as needed, consecutively from left to right, as we see them on the screen. If a line segment contains more than two points (recall that these are discrete), then they will be labeled with only their endpoints. Each "curve" in a list will be separated by a comma. The ordering in the list is irrelevant. The topic in this post involves the unordered listing and number of the total of discrete "curves" in two dimensions for n distinct points, where n is greater than or equal to one. Examples: -------------- n = 1 <----------------- A ----------------> List: {A} Total: 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - n = 2 <-------------- A ----------- B -------------> List: {A}, {B}, {A, B}, {AB} Total: 4 Note: Here, you cannot have {A, AB}, {B, AB}, or {A, B, AB}, because neither point A nor point B is isolated once you identify AB as a line segment in the set. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - n = 3 <-------- A -------- B -------------- C -----------> List of points only: {A}, {B}, {C}, {A, B}, {A, C}, {B, C}, {A, B, C} List of line segments only: {AB}, {BC}, {AC} List of points and line segments mixed: {A, BC}, {AB, C} Total: 12 Note: Here, you cannot have {B, AC}, because point B is not an isolated point. It is part of line segment AC, and AC has already been listed. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Puzzle: Compute the total number of "curves" for n = 4. *** Optional additional question: If you continue on getting totals for n = 5, 6, 7, 8, ..., you will get a sequence that might suggest a pattern to you. As with all sequences, there are an infinite number of pos- sibilities for a formula. However, there is a formula that I have in mind. Whatever formula you get, do not attempt to prove it. Just type the formula in your reply.

DejMar, this is getting away from me. I intended the factorial symbol to just be used as a factorial, not also as a subfactorial. Also, on another site, a user stated that the double factorial is one character, so I don't know if that should still be considered.

My computer isn't letting me put post # 8 in a quote box. Let me leave out the word "unless" for my intended meaning. "Slight modification of the rules: An expression isn't valid if at least one operation, or set of parentheses, or both can be removed, and the expression is still equal to 24."

witzar, first one -- I know you stated to ignore the ones with redundant parentheses. However, all parentheses have to be left off this one, because "8 + 8 + 8" manages on its own to equal 24. Modified rule. ------------------------------------------------------- fourth one -- The inside factorial sign would be left off, because the expression that is equal to 2, is already equal to 2 factorial. Modified rule. ** It's a variation/repeat of the third one. ------------------------------------------------------ sixth one -- ** See the immediate comment above. It's a variation/repeat of the fifth one.