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There are  18 chairs numbered from 1  through 18 around a circular table. How many ways can three people be seated, so that no two people are adjacent?

 Jan 20, 2020
 #1
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We can use a stars-and-bars technique.  First, we cut the circle at a certain point, to get a row of chairs.  Using stars-and-bars, there are C(16,3) = 560 ways to choose two chairs so that no two chairs are adjacent.  Therefore, there are 560*6 = 3360 ways to seat the three people.

 Jan 20, 2020
 #2
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I tried this before an got in a mess. This time I am more confident.

But there is still plenty of room for error.

 

Seat each person and tie their chair to the chair on the right (which will be empty.)

Think of the 2 chairs and the one person as a single unit.

So now you have 3 fat units and 12 more chairs.

 

You have to slip the fat units between the other chairs.

15 units altogther and you need to choose the 3 that are fat.

so that is 15C3= 455

Now the three people can be in any order so that is  3!=6  ways

 

455*6=2730 ways

 Jan 21, 2020
 #3
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I tried both 3360 and 2730, and both are incorrect. Help!

 Jan 22, 2020
edited by ramenmaster28  Jan 22, 2020
 #4
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Do you know what answer they (the asker) believe to be correct?

 

When you get the answer can you share it with us please. 

Also the method.

Melody  Jan 22, 2020
edited by Melody  Jan 22, 2020
 #5
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What abuit if you put the first fat person anywhere.  15 places.

then there are 14 places to choose from and 2 have to be fat people

15*14C2=1365

 

1365*2=2730

 

Slightly different method but still I get the same result.

(The logic is not very different)

 Jan 22, 2020
 #6
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The first fat person? Did you purposely type that?

Guest Jan 22, 2020
 #7
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Sort of, it is not a typo or a slir, he is fat becasue he is sitting on a chair and has a second chair tied to him on his right. 

Earlier I referred to this (one man 2chairs) as a fat unit.

Melody  Jan 22, 2020

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