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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 8, 2020
 #1
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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.

 Jun 8, 2020
 #2
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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

 Jun 9, 2020
edited by Forgo  Jun 9, 2020

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