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Interesting Question

In the Samsung Galaxy Tab, you can put a passcode to protect it. There are 9 dots on the passcode. You have to connect at least 4 dots to make a combination password. How many passcodes can you make? Details: 1. If you try to connect one corner to another corner, it counts as THREE dots. This is because there is a dot in between. 2. Order DOES matter. 3. Have fun! This is a problem I have thought about for a few days, but have no idea where to start.

Note by Anton Than Trong
3 years, 10 months ago

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This is a great example of interesting questions which students submit that get rejected because they aren't clearly phrased. You have an opinion of what can be done, but that has not been conveyed well which results in some confusion.

For example, you have not provided the basic explanation of what can, or cannot be connected. For example, can I connect up $$1-3-2-5$$? I believe that your answer is no, but I do not see which aspect this will contradict.
Can dots be reconnected again - i.e. is $$1-2-3-2$$ valid?
What does Condition 1 really mean? It is badly phrased, and I can think of several different interpretations of it.

As always, it is important to express yourself clearly, so that you can easily be understood by others, who would then be able to give you better feedback about how to approach the problem. Staff · 3 years, 10 months ago

I understand what he's trying to say: The 9 dots are arranged in a 3x3 grid. A passcode is formed by joining at least 4 of these dots with lines. You can NOT choose a dot more than one time, however it is possible to draw a line passing over an already selected dot, but then it won't be registered again. Condition 1 means that you can't draw a line over a dot without selecting it. So if you draw a line from e.g. the bottom left corner to the upper left corner (and the dot in between is not already selected), you will select the point in between them as well. Any questions? · 3 years, 10 months ago

The problem implicitly relies on the assumption that one understands the workings of the passcode on the Galaxy Tab. The issue is that you are reading more into the question than is currently stated.

1. You cannot choose a dot more than one time.
2. The path is formed using straight lines. This implies that we can't connect 1 to 9 without also choosing 5. 3. It is possible to draw a line over a chosen dot.

Is this set of additional constraints a necessary and sufficient set? Are there any other conditions that are missing? Any other edge cases that we need to consider? Staff · 3 years, 10 months ago

This is the list of full and complete restraints of the problem. · 3 years, 10 months ago

Yes, that's what I realized too. The start and end dots can be chosen however you like. Apart from that, I believe that's sufficient. · 3 years, 10 months ago

4 dots: 1624 solutions · 3 years, 10 months ago

its not a straightforward permutations question, the answer changes based on the choices you make for the 1st, 2nd and 3rd dot. · 3 years, 10 months ago

but its written atleast 4 dots means more can be possible · 3 years, 10 months ago

by the way when you join two corners there are 3 dots involved, so why does it count as two? · 3 years, 10 months ago

It connects the corner spots directly, passing over the middle one. Got? · 3 years, 10 months ago

Yeah but 3 dots are involveD? · 3 years, 10 months ago

16 · 3 years, 1 month ago

9C4 ways .There are 9 dots,while each combination takes up at least 4 dots. · 3 years, 10 months ago

No, it is not that straight forward. There are restrictions on how the patterns are formed. · 3 years, 10 months ago

Thank you for the correction. · 3 years, 10 months ago

No problem. · 3 years, 10 months ago

362844 · 3 years, 10 months ago

HOW? · 3 years, 10 months ago

did u took the cases of selecting 4 and more dots out of 9 and adding them? · 3 years, 10 months ago

and permuting would exceed his answer. · 3 years, 10 months ago

That would never yield that many results. · 3 years, 10 months ago

may be possible......but sooo big digit answer.......??????!@ · 3 years, 10 months ago

I already tried that logic it would reach around 700,000. · 3 years, 10 months ago

ok.....but how to begin with this question? · 3 years, 10 months ago