Probability of $r$ people in between The question is as follows, I will append my solution as well. I have a suspicious feeling that my approach is utterly wrong.

Mike and Tom are among $n$ people who are arranged at random in a line. What is the probability that exactly $r$ people stand between them?

My solution:
$$A= \{\text{Exactly }r\text{ people stand between Mike and Tom}\}$$
So $n(S)=(n+2)!$ where $n+2$ stands for Mike, Tom and the $n$ other people.
$$n(A)=2\binom nr(n-r)!r!$$
Hence the probability is
$$P(A)=\frac{2\binom{n}{r}(n-r)!r!}{(n+2)!}$$
Is there any problem with my solution? I actually worry about the numerator part.
Edit: Edited the question from the previous clarification edit. Question actually includes Mike and Tom in n people.
 A: Suppose there are $n$ people in total and consider the picture:
$$
\underbrace{*\cdots *}_k\textrm{M}\underbrace{*\cdots*}_r\textrm{T}\underbrace{*\cdots *}_{n-r-k-2}
$$
for $0 \leq k \leq n - r - 2$. There are $n - r - 1$ possible locations for Mike. Flipping the location of Mike and Tom gives $2(n  -r - 1)$ possibilities. There are $(n - 2)!$ possibilities for the remaining people. Hence, there are $2(n - r - 1)(n - 2)!$ possible configurations. Thus the probability is
$$
\frac{2(n-r-1)(n-2)!}{n!} = \frac{2(n - r- 1)}{n(n-1)}
$$
A: The number of ways to arrange just Mike and Tom is $^n\!P_2=n(n-1)$. If there are $r$ people between them ($0\le r\le n-2$) there are $n-r-1$ ways to pick where the first of them will go, then two ways to pick who goes into that first slot. In both cases the number of ways to arrange the other people remains the same, so the answer is
$$\frac{2(n-r-1)}{n(n-1)}$$
Your approach fails to consider the $r$ in-between people being able to occupy any part of the line, which generates the $n-r-1$ factor.
