Permutations / Probability sitting around a table James and his 5 friends(Kyle,Tom,Bob,Michael,John) are about to sit around a round table there are 6 sits at the table.
What is the probability James will sit between Bob and Michael ?
My solution
A= number of ways James can sit between Bob and Michael
B= total number of ways they can sit
As James, Bob and Michael technically sit next together we can consider them as single unit
A=4!* 2! The 2! Is the number of ways Bob and Michael can swap.
B= Total number of ways the can all sit is (n-1)!
I.e 5!
P= A//B
Which is 2/5
 A: That is not the right answer. Instead break the circle at James so the remaining five people form a line; there are $2\cdot3!$ seating arrangements out of $5!$ where Bob and Michael sit at the line's ends, and hence next to James in the full circle (the other three people may be permuted arbitrarily). Thus the answer is $\frac{2\cdot3!}{5!}=\frac1{10}$.
A: By convention, in a circular permutation,  only the relative order of the objects matters.
Seat James.  We will use him as our reference point.  Relative to James, there are indeed $5!$ ways to arrange his friends around the table as we proceed clockwise around the table from James.
For the favorable cases, once James is seated, there are $2$ ways to arrange Bob and Michael next to James.  That leaves three seats for Kyle, Tom, and John.  They can be seated in those three seats in $3!$ ways as we proceed clockwise around the table from James.  Hence, the number of favorable cases is $2 \cdot 3!$.
Therefore, the probability that James will sit between Bob and Michael if the six people are seated randomly around the table is
$$\frac{2 \cdot 3!}{5!}$$
In your attempt, you forgot to use a reference point for the favorable cases.
A: I am assuming that you mean that the three people are ordered [Bob, James, Michael] or [Michael, James, Bob].
You can express the probability as
$$\frac{N}{D},$$
where $D = (5!)$.
That is, since the table is round, you can arbitrarily designate anyone as the head of the table (i.e. the twelve-o'clock position).  Then you will have $(5)$ remaining people to permute.
So, based on the above discussion, the problem reduces to computing $N$.
Think of the $(3)$ people, Michael, James, Bob, as forming one fused unit.  Then, in effect, you have $(4)$ units to permute around the table.  Namely, the fused unit and the other $(3)$ units.
Since the table is round, this can be done in $(4-1)!$ ways.
Further, the fused unit permits $(2!)$ internal permutations of the three people inside the unit, since James must be in the middle.
Putting this all together, the probability is
$$\frac{[(4-1)!] \times 2!}{(6-1)!}.$$
An alternative way of computing $N$ is to arbitrarily decide that James will go in the twelve-o'clock position.  Then, Bob, Micheal can be permuted in $(2)!$ ways around James, and then the $(3)$ remaining people can be permuted in $(3!)$ ways in the three remaining seats.
