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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.

 

I did this.

1. There are 7 choices in the heptagon for N to be labeled. 
2. After the N has been labeled, there are now 4 remaining spots for one O to go.
3.  Then label the other O -- there are 3 choices now, we subtract a choice for where the previous O was placed.
4. With all the remaining letters, there are 4! ways to label the rest of the heptagon. 
5. We then take all the information and multiply it --  7*4*3*4! = 2016
6. After this is done we need to divide 2016/7 to account for the rotations of the heptagon. This equals 288.
7. The last step is to correct for over counting the O's and to divide by 2!
The answer to the problem is, there are 144 distinct ways to label the vertices of the heptagon. 

 

IS it right if not please explain right answer and why

 Jun 21, 2020
 #1
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These questions can always be looked at in a number of different ways.

I got the same answer as you, I think your answers are fine.

 

Here is my logic (possibly the same as yours)    But definitely more simply displayed.

 

 

1) Put down the N

2) 4 places available for the two Os so that is   4C2 = 6 ways

3) 4 distinct letters left for 4 distinct places so that is 4! = 24 ways.

 

6*24 = 144 possible ways     (where rotations are classed as the same)

 Jun 22, 2020
edited by Melody  Jun 22, 2020
 #2
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The only reason that I saw this question is that it had been automatically flagged.

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 Jun 22, 2020

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