+232 Lance has a regular heptagon (7 sided figure). How many distinct ways can he label the vertices of the heptagon with the letters in OCTAGON if the N cannot be next to an O? Rotations of the same labeling are considered equivalent.
+118673
Lance has a regular heptagon (7 sided figure). How many distinct ways can he label the vertices of the heptagon with the letters in OCTAGON if the N cannot be next to an O? Rotations of the same labeling are considered equivalent.
OCTAGN O
6!/2 = 360 (i think) , that is with no restrictions on N (so the answer must be less than this.
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After thought I have decided to tackle this by placing the 2 Os into position before adding the other letters.
Each time I will let one of the 0 dictate the starting point of the circe.
The second O can then be placed in the 2nd, 3rd or 4th positon.
I will takle each of these scenarios seperately.
\(OO \overline{N}***\overline{N} \qquad 1*4*3*3!=72\\~\\ O\overline{N}O \overline{N}**\overline{N} \qquad 1*4*3*2*2! = 48 \\~\\ O\overline{N}\overline{N}O\overline{N}\;N\;\overline{N}\qquad 1*4*3*2*1*1=24\\~\\ O***O**\quad \text{This is the same as the last one.}\)
72+48+24= 12(6+4+2) = 12*12 = 144
So I get 144 but there is a lot of room for errors.
coding:
OO \overline{N}***\overline{N} \qquad 1*4*3*3!=72\\~\\
O\overline{N}O \overline{N}**\overline{N} \qquad 1*4*3*2*2! = 48 \\~\\
O\overline{N}\overline{N}O\overline{N}\;N\;\overline{N}\qquad 1*4*3*2*1*1=24\\~\\
O***O** \text{This is the same as the last one.}
+118673 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 
