Cyclic permutations: Ten persons sit around a circular table

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Ten persons -- A, B, C, ..., I, and J -- sit around a circular table. The chairs are all numbered.

What is the probability that A and B sit next to each other?
 
Cyclic permutations. I hope I've got this right. I trust pka will straighten me out, if not.

There are (n-1)! ways to arrange n people around a circular table.

Now, A and B must be seated next to one another. Think of them as tied together or as 1 morbidly obese person. Therefore, you have 8! possible arrangements. Also, the 2 can be arranged in 2! ways among themselves.

So, (8!)(2)=80640

The number of possible arrangements around the table are 9!=362880

80640/362880=2/9
 
Re: Cyclic

Hello, atomos!

I have a back-door approach to this one . . .

Ten people {A, B, C, ⋯ ,J}\displaystyle \{A,\,B,\,C,\,\cdots\,,J\} sit around a circular table.
The chairs are all numbered.
What is the probability that A\displaystyle A and B\displaystyle B sit next to each other?
Click to expand...

A\displaystyle A can sit in any of the ten chairs . . . right?

Then the probability that B\displaystyle B sits in an adjacent chair is: \(\displaystyle \L\,\frac{2}{9}\)

 
Hello, Cody!

Yes . . . and good work!

But when I saw your simple answer (29)\displaystyle \left(\frac{2}{9}\right), I just had to see why.

And when I saw why, it was AH-HA! time
    \displaystyle \;\;(My big chance to look like a genius . . . )

 
not circular permutation?

I think itsn't a circular permutation, since all chairs are numbered. Eventhough the table is round.
So I'm sorry if I said it was circular permutation :wink:

so the probability of A and B sit next to each other :
10.8!10!=19\displaystyle \frac{10.8!}{10!}=\frac{1}{9}
Is it right or wrong ?
thanx
 
not circular permutation?

I think itsn't a circular permutation, since all chairs are numbered. Eventhough the table is round.
So I'm sorry if I said it was circular permutation :wink:

so the probability of A and B sit next to each other :
10.8!10!=19\displaystyle \frac{10.8!}{10!}=\frac{1}{9}
Is it right or wrong ?
thanx
 
What possible difference could the fact that the chairs are numbered make?
Seat A at the table. Once that is done the table is ordered.
There are nine places to seat B, two of them are next to A: 2/9.
 
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