Probability that Ken and John set next to each other
Solution 1:
Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.
Solution 2:
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 $$P(A)=\frac{2!8!}{9!}=\frac{2}{9}$$