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.
There are 7! = 5040 ways to label the heptagon. There are 4! = 24 ways where the O's are next to an N, so the number of ways where N is not next to an O is 5040 - 24 = 5016.
BTW, That's incorrect.
Label one vertex of the heptagon as any variable you choose.
The O's can't be next to the point N.
There are \(\binom42 = 6\) ways to choose the 2 spots. Since there are 4! ways to label the other vertices, we multiply 4! = 24 by 6 to get 144.
DO NOT COPY THIS SOLUTION WORD FOR WORD.
I know this is an AoPs problem but I also rlly needed help so.. Yeah.
Make sure to understand method used to solve this problem