Probability and arithmetic sequence question Three numbers are chosen at random from $1,2,...2n$ with $n > 1$. Show that the probability that the numbers are in
A.P. is $\frac{3}{4n-2}$
I don't know how to deal with this. Please help.
 A: The total number of $3$-number combinations from $1$ to $2n$ is $\binom{2n}{3}$. Now consider $1$ as the first taken number. Then, the $3$-number A.P. with the greatest common difference will occur if the then highest chosen number is $2n-1$ as common difference is $\dfrac{(2n-1)-(1)}{2}=n-1$. Hence the combination of numbers in A.P. with greatest common difference and with $1$ chosen as one of the numbers is $\{1,n,2n-1\}$. Now we start to narrow our common difference until it becomes one and notice that with $1$ as the first chosen number, there are a total of $n-1$ combinations of $3$ numbers forming an A.P. Repeat the process for $2$ and you'll notice that the number of combinations of $3$ numbers forming an A.P. is again $n-1$. However for $3$ it becomes $n-2$. For any $k$ chosen as the first number, there are $n-1-\left\lfloor \frac{k-1}{2}\right\rfloor$ such possible combinations. Sum it all up to $2n-2$ (since the last choice for $k$ is $2n-2$ for which there is only one A.P. combination, namely $\{2n-2,2n-1,2n\}$), you get $\displaystyle\sum_{k=1}^{2n-2}n-1-\left\lfloor \frac{k-1}{2}\right\rfloor=n(n-1)$.
Then
\begin{align}
Pr&=\frac{n(n-1)}{\binom{2n}{3}}\\&=\frac{n(n-1)}{\frac{2n!}{3!(2n-3)!}}\\&=\frac{3!n(n-1)}{2n(2n-1)(2n-2)}\\&=\frac{3}{4n-2}
\end{align}
