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# Counting and Probability Question

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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?

Dec 29, 2019
edited by mathviolinjk  Dec 29, 2019
edited by mathviolinjk  Dec 29, 2019

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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.

Dec 29, 2019
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It is interesting that you can answer a question when half of it is missing.

Melody  Dec 29, 2019
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I think that Guest noticed that this question looked similar to the link you posted, so he assumed it was the same one, since both solutions are the same.

Guest Dec 31, 2019
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Try proofing it before you post!

Dec 29, 2019
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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
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