A combinatorics problem with positions 
Thomas and Peter are in a team of $n$ people that are ordered in a line. What is the probability that there are exactly $k$ people in between them?

My take:
There are $n!$ ways the $n$ people can be ordered.
From the $n$ positions we choose $2$ for Thomas and Peter.
 We multiply this by $2$, because Thomas and Peter could exchange positions. From the remaining $n-2$ positions we choose how many there are in between Thomas and Peter. And the remaining $n-2$ people can be ordered in $(n-2)!$ ways. So the required probability is 
$$P(A)=\frac {\binom {n}{2}2\binom {n-2}{k}(n-2)!} {n!}=\binom {n-2}{k}$$
But the answer given is $\frac {2(n-k-1)}{n(n-1)}$.
Any help appreciated.
 A: If we number the positions $1$ through $n$, with $1$ being the 'leftmost' point of the line, then the leftmost of Peter and Thomas can take positions $1$ through $n-(k+1) = n-k-1$, because the leftmost still needs $k$ persons and the rightmost of the two after that.
Given that the leftmost can be either Peter or Thomas, and given that the remaining $n-2$ people can be arranged in $(n-2)!$ ways, this gives us $2(n-k-1)(n-2)!$ possible line-ups with $k$ people between Peter and Thomas.
Given that there are $n!$ possible lineups, the probability is:
$$\frac{2(n-k-1)(n-2)!}{n!} = \frac{2(n-k-1)}{n(n-1)}$$ 
A: Here is one way to think about it. Say Thomas comes before Peter; then once we place Thomas, say at spot $a$, then Peter must come at spot $a+k+1$. Note that we can only do this if $1\leq a\leq n-k-1$, since we only have $n$ spots to place the people in. Once we place Thomas and Peter, we may arrange the remaining $n-2$ people arbitrarily in the remaining spots; there are $(n-2)!$ ways to do this. So there are $(n-k-1)\cdot (n-2)!$ ways in which Thomas can go first, and then multiply by $2$ to account for when Peter comes first for a total of $2\cdot(n-k-1)\cdot(n-2)!$ desirable cases. As you said, there are a total of $n!$ ways that the people may be arranged, so the probability is 
$$
\frac{2\cdot(n-k-1)\cdot(n-2)!}{n!}=\dots
$$
A: Treat Thomas, Peter and the $k$ members as one entity. No of ways to arrange them will be $(n-k-1)!$. Now in this group, Thomas and Peter are on extreme end. They can interchange in $2!$ ways. Remaining $k$ people can be chosen from $n-2$ (ie excluding Thomas and Peter) people in $\binom{n-2}{k}$ ways and can interchange position in $k!$ ways. So in all we have:
$$P = \frac{2(n-k-1)! \,\binom{n-2}{k}\, k!}{n!}$$
which on simplification leads to
$$P = \frac{2(n-k-1)}{n(n-1)}$$
A: The number of ways to choose the 2 positions for Tom and Peter is not $\binom{n}2$, because this does not incorporate the condition that Tom and Peter  must have $k$ spots between each other. The actual number is smaller.
Think about this: how many ways are there to choose the leftmost of the two spots? The answer is less than n, since this spot cannot be too close to the right edge. Once you've chosen the leftmost spot, there is now a unique choice for the rightmost spot.
A: There are $\dbinom{n}{2}$ ways to choose the 2 places for Tom and Peter in the line; 
and if there are exactly $k$ people between them, then $a+b=n-k-2$ 
where $a$ is the number of people to the left of both of them and $b$ is the number of people to their right. 
Therefore the probability is given by $\displaystyle \frac{n-k-1}{\binom{n}{2}}=\frac{2(n-k-1)}{n(n-1)}$.
A: $\underline{Another\; Way\;}$
Place $A$ anywhere on a circle with $n$ seats, there are $2$ ways to place $B\;$ with $k$ seats in between,
and the circle can now be "straightened" by cutting at any of $n-k-1$ points between seats
outside the block $A\; o\; o\; o\; o\; o\; B$
Unconstrained, $A\;$and $B$ can be seated in $n(n-1)$ ways, thus $Pr = \dfrac {2(n-k-1)}{n(n-1)}$
