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.

The way that I did this problem is to first label the N - there would be 7 choices for where to label it. Then label one of the Os - there would be 4 choices for where to put it. Then label the other O - there would be 3 choices (-1 for where we put the previous O). Then label all the other letters - 4! ways to do that. Then, divide by 7 to account for rotations of the same labeling.Doing that, I got 288. Is this right? Or where did I make a mistake?

Thanks in advance!

mathviolinjk Dec 29, 2019

#1**0 **

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.

Guest Dec 29, 2019

#2

#4**-4 **

Sorry about that, I did not realize that you could not see parts of my question (I could see all of it). I tried re-typing my question; hopefully you can see it now. More specifically, I was wondering where I went wrong in my solution. Thanks!

mathviolinjk
Dec 29, 2019