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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.

 Jun 25, 2020
 #1
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^(add on to my first post) I found that there are 7!/2! ways to label the heptagon considering that there are two O's in "OCTAGON". Since there are 4! = 24 ways where the O's are next to an N, so is the answer 7!/2! - 24?

 

Am I on the right track?

 Jun 25, 2020
 #2
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I answered this a little while ago.

It is normal to assume that rotations are the same.

 

Put the N anywhere.

now there are 4 places where the os can go.    4C2 = 6

Now there is 4 spots not taken and 4 letters to put into them, that is 4! ways.

 

So the answer is 6*4! = 144

 Jun 25, 2020
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Why don't you multiply 144*7 at the end of the problem to consider the $N$

 Jun 26, 2020
 #4
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Because in these problems it is usual to consider rotations as identical.

So one of the elements can go anywhere. Then that becomes the 'beginning' of the loop

Melody  Jun 27, 2020
edited by Melody  Jun 27, 2020

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