Question: Find $a$ and $b$ so $ h(x) = anx^n - b(n + 1)x^{n-1} + x + 2 $ is divisible by $ x^2 -3x+2$.

I know $ h(1) = h(2) = 0 $ but what do I do with $n$

Answer: 
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

$$
\frac{2(n-r-1)(n-2)!}{n!} = \frac{2(n - r- 1)}{n(n-1)}
$$
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.