Pay no attention to guests nonsense answer. I will hide it as it adds to confussion.
I have answered this on the original thread.
https://web2.0calc.com/questions/tricky-question_8#r3
Think about it hard but then if you have more questions you can ask.
If I do not give a timely response it might be becasue i have not seen your question.
If you were a member you could send me a private message with a link.
But since you are not you might have to be more creative.
A guest has asked me to try and explain my answer better.
This is the request:
https://web2.0calc.com/questions/i-do-not-understand
Hi interested guest,
For starters
O stands for the letter O
N stands for the letter N
\(\bar N\) is a a common set notation, it means NOT N
\(*\) means any available letter. A wild card if you like.
So where I have put \(\bar N\) I mean it can be any of the letters available EXCEPT for N
Since there is 7 letters in OCTOGON each of my strings has the seven letters.
Remember that they are really in a circle so the last on joins to the first one.
N cannot be next to an O so I worked out that this could be looked at in 3 distinct scnearios.
These are the 3 strings I have used.
When you put items in a circle and rotations are counted as the same you deal with that by placing one of the objects first.
It makes no difference which one or where.
1) In my case I chose to place an O first, so each of my strings begin with O.
2) then I added the possible postions for the other O . I found three distinct possibilities.
3) Then for each one I added in where N could NOT go.
4) The remaining places could be filled with ANY available letter.
5) After I worked out what stings would work. I then worked out how many permutations would be possible for each one.
If you want me to go through that i can but I want you to understand all the rest I have talked about first. So if I am to offer more explanation you will need to interact with me further.
I hope you can work it out. It is really nice that you have shown a genuine interest