calculating probability, people in a row There are total of n people, among whom A and B, stand in a row.
Q1: P(exactly r people between A and B)=?
Q2: Instead of in a row, they stand in a ring. 
    P(exactly r people between A and B)=?
For Q1, I am only this far, which is not too far..(LOL): $\binom{n-2}{r}\times r!$ for the combination for people between A and B. (......A,r people,B...).But I don't know how to handle the combination for the two sides, to the left of A and to the right of B.
For Q2, honestly I cannot tell how the combination could be different but I know they should be different. 
Thanks for your help!!
 A: $1)$ Assume $A$ stands ahead of $B$. By symmetry you then multiply by $2$. $A$ can be in positions $1,2,...,n-r-1$. The remaining can be configured in $(n-2)!$ ways. So the probability is $$\dfrac{2(n-r-1)(n-2)!}{n!}$$
$2)$ Call some position in the ring slot $1$ and number them clockwise. No matter where $A$ is, there can be $r$ people between $A$ and $B$ in two different ways, unless $r=n/2-1$. They are slots $A+r+1$ and $A+n-(r+1)$. Further, $A$ can occupy slots $1$ through $n$.
So if $r\neq n/2-1$ the probability is
$$\dfrac{2n(n-2)!}{n!} = \dfrac{2}{n-1}$$
else
$$\dfrac{n(n-2)!}{n!}=\dfrac{1}{n-1}$$
A: Maybe a bit more straightforward line of thinking about the first part is to count first in how many ways a group of adjoining elements of size $r$ can be placed in the given row. Clearly, none of elements of the group can't take first or last place of the row (otherwise we couldn't fit $A$ or $B$) so there are only $n-2$ places left for use. Considering some particular cases we can quickly conclude that the block of $r$ elements can be placed in $(n-2)-r+1 = n-r-1$ ways. The group of r elements that will sit between A and B is simply an $r$-permutation of set $n$ without A and B, i.e. those men can be chosen in $(n-2)_r$ ways from $n-2$. The A and B encloses this group in 2 ways (A from left, B from right and vice versa). The rest of $n-r-2$ elements can fill rest of the places in $(n-r-2)!$ ways. The sample space is clearly all possible permutations of set $n$, which size is $n!$.
Hence,
$$p = \frac{2(n-r-1)(n-2)_r(n-r-2)!}{n!} = \frac{2(n-2)_r(n-r-1)!}{n!}$$
Regarding second part, it may be useful to enumerate the circle and consider sample space as all possible tuples (place of $A$, place of $B$) which size is $2$-permutations from set n which is equal to $(n)_2$. Brief inspection reveals that no matter how many elements there are between $A$ and $B$, the group can be rotated $n$ times to produce different positions. Taking into account relative position of A against B, the number of ways to place $A$ and $B$ in the circle to have $r$ elements between them is 2n.
Hence,
$$ p = \frac{2n}{(n)_2} = \frac{2}{n-1}$$
A: Is it okay if I take the denominator as $^{n}P_{2}.$ My logic is that we are essentially searching for the permutations that $A$ and $B$ can have in a circular ring of $n$ people. The numerator would be $2n,$ because for each n ways in which $A$ can be arranged, $B$ can be arranged in two ways - giving us a total of $2n.$ 
So the answer becomes  $\frac{2n}{^{n}P_{2}}.$
The answer gives me $\frac{2}{n-1} ...$ but I wanted to know if the thought pattern was right ?. If not, where am I going wrong!
