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In how many ways can 8 people sit around a round table if Pierre and Thomas want to sit together, but Rosa doesn't want to sit next to either of them? (Treat rotations as not distinct but reflections as distinct.)

Guest Mar 19, 2018
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 #1
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Pierre and Thomas can be "anchored" together....and they can be seated in 2 ways

 

Rosa  can sit in any of the 4 positions not next to Pierre and Thomas

 

And the other 5 people can be seated in 5!  =  120 ways

 

So.....the possible seating arrangements are

 

2 * 4 * 120  =   

 

960

 

 

cool cool cool

CPhill  Mar 19, 2018
 #2
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Sorry, I don't get it !!. See the exact question and the three "answers" here and how different they are from each other. This shows the difficulty involved in probability, even though this one is a "counting problem"!!

https://answers.yahoo.com/question/index?qid=20150403142524AAJS3T0

Guest Mar 19, 2018
edited by Guest  Mar 19, 2018
edited by Guest  Mar 19, 2018
 #3
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My understanding of this is as follows:

There are 8 people, 2 of which must sit together. So, we have 6 + 1(as one unit) =7

The 2 people can sit together in 2! ways. The remaining 6 can be seated in 6! ways without any restrictions. 

So we have a total of: 2! x 6! =1,440 ways without any restrictions. But, we have a restriction that Rosa cannot sit on either side of Pierre and Thomas. Of the 6! ways that the 6 people can sit, Rosa sits to the left of Pierre and Thomas in 6!/6 = 120 ways. And, similarly, she can sit to the right of Pierre and Thomas in 6!/6 = 120 ways. So, 120 x 2 = 240 positions that must be excluded from 1,440. Or:

1,440 - 240 =1,200 ways that the 8 people can be seated on a round table with the restrictions given.

Guest Mar 19, 2018
 #4
avatar+969 
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This type of combination problem is elementary. There is nothing difficult about it once you have mastered basic combination counting skills.  For you, that will never happen!!

 

Another thing this problem demonstrates is the difficulty in teaching blarney bankers, curmudgeons, and others with hardened arteries, atrophied brains, and general defects in basic intellect, present since birth or, in your case, hatching.    

 

Don’t you have anything better to do, like counting Monopoly money? 

 

 

GA

GingerAle  Mar 19, 2018
 #5
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Solution 1: We choose any seat for Pierre, and then seat everyone else relative to Pierre. There are 2 choices for Thomas; to the right or left of Pierre. Then, there are 4 possible seats for Rosa that aren't adjacent to Pierre or Thomas. The five remaining people can be arranged in any of 5! ways, so there are a total of 2 x 4 x 5! = [960] valid ways to arrange the people around the table.

Solution 2: The total number of ways in which Pierre and Thomas sit together is 6! times 2 = 1440. The number of ways in which Pierre and Thomas sit together and Rosa sits next to one of them is 5! times 2 times 2 = 480. So the answer is the difference 1440 - 480 = [960].

Guest Mar 19, 2018

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