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.}
