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?
Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.
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 $$P(A)=\frac{2!8!}{9!}=\frac{2}{9}$$
