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A group of ten people sits down, uniformly at random, around a table. Ken and John are part of this group. Determine the probability that Ken and John sit next to each other.

There are $10!$ ways to arrange the seating for everyone, there are 10 possible ways for John and Ken to sit together.

$$\operatorname{Pr}(J\ \&\ K ) = \frac{10}{10!} = \frac{1}{9!}$$

Am I correct?

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    $\begingroup$ There would be $10!$ ways to arrange the seating if they were in a line. (Not on a circle.) $\endgroup$ – amcalde Nov 11 '14 at 2:38
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    $\begingroup$ $10/10! = 10/9!$ - what? $\endgroup$ – Suzu Hirose Nov 11 '14 at 3:31
  • $\begingroup$ @SuzuHirose Hilarious typo! $\endgroup$ – lzc Nov 17 '14 at 14:20
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Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.

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    $\begingroup$ @Micah: Thanks for making the John/Ken fix. I missed that one (obviously). $\endgroup$ – paw88789 Nov 11 '14 at 10:24
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Please take a look at Circular Permutations

Total possible arrangements $$(10-1)!=9!$$

Consider $J$ & $K$ as one unit Now we'll have $9$ units then but (internal) arrangements of those two is not considered so favorable arrangements are $$2!(9-1)!=2!8!$$

Hence $$P(A)=\frac{2!8!}{9!}=\frac{2}{9}$$

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    $\begingroup$ Thanks for pointing out circular permutations for me. I had no idea! $\endgroup$ – lzc Nov 11 '14 at 2:41

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