# Probability that Ken and John set next to each other

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?

• There would be $10!$ ways to arrange the seating if they were in a line. (Not on a circle.) Nov 11, 2014 at 2:38
• $10/10! = 10/9!$ - what? Nov 11, 2014 at 3:31
• @SuzuHirose Hilarious typo!
– lzc
Nov 17, 2014 at 14:20

Two of nine people sit next to John. The probability that Ken is one of these two is $\frac29$.

• @Micah: Thanks for making the John/Ken fix. I missed that one (obviously). Nov 11, 2014 at 10:24

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

• Thanks for pointing out circular permutations for me. I had no idea!
– lzc
Nov 11, 2014 at 2:41