There are many approaches to an answer. Note that the job cannot be done if $n$ is odd. So let $n$ be even, say $n=2m$.
For concreteness, let $n=10$.
Line up the students in increasing order of student number. The first person in the list can choose her study partner in $9$ ways. For every such choice, the so far unpartnered student with smallest student number can choose her study partner in $7$ ways.
For every such choice, the so far unpartnered student with smallest student number can choose her study partner in $5$ ways. For every such choice, the so far unpartnered student with smallest student number can choose her study partner in $3$ ways. That gives a total of $(9)(7)(5)(3)$ ways.
The answer looks a little nicer if we write instead $(9)(7)(5)(3)(1)$.
Maybe to make things look even nicer, we can multiply top, and missing bottom, by $(10)(8)(6)(4)(2)$. We get
$$\frac{10!}{(10)(8)(6)(4)(2)},\quad\text{which is} \frac{10!}{2^5 \cdot 5!}.$$
Now that we have gone through the reasoning got $n=10$, the general case is straightforward. Suppose there are $2m$ students.
The student with lowest student number can choose her partner in $2m-1$ ways. For every such way, the so far unpartnered student with lowest student number can choose her partner in $2m-3$ ways. And then the so far unpartnered student with lowest student number can choose her partner in $2m-5$ ways, and so on, for a total of
$$(2m-1)(2m-3)(2m-5)\cdots(3)(1).$$
The same manipulation as before gives the more compact expression $$\frac{(2m)!}{2^m \cdot m!}.$$
Let $f(n)$ be the number of ways to sort the $3n$ students into $n$ trios, with the condition imposed.
- Suppose all three students in grade $n$ are together. There is $1$ way to put them together, and then $f(n-1)$ way to put everyone else into trios.
- Suppose two students in grade $n$ are together. The third student must be in grade $n-1$. That means that there is another trio containing one student in grade $n$ and two students in grade $n-1$. There are $9$ ways to arrange this, and then $f(n-2)$ ways to put the younger students into trios.
Therefore we have $$f(n)=f(n-1)+9f(n-2)$$ with the starting values $f(1)=1,\ f(2)=10$.
This is equivalent to Emperor of Ice Cream's second solution. The other solutions are correct if we are looking only for the number of schemes (e.g. 111-223-233-445-455). The answer to the original question is $$f(5)=280.$$.
Best Answer
Please note that ${2n \choose n} $ orders the groups so you need to divide it by $2!$ as you only want to find number of ways to make two groups of $n$.
So the probability using your method is,
$ \displaystyle 1 - 2! \cdot {2n-2 \choose n-1} / {2n \choose n} = \frac{n-1}{2n-1}$
The easiest way to find the probability is to first place $A$ in any group. Now that leaves $(2n-1)$ places across two groups for $B$ but if $B$ has to be in the same group as $A$ then it must be in one of the $(n-1)$ places in $A$'s group. That is probability of $ \displaystyle \frac{n-1}{2n-1}$.
A bit more explanation:
Say I have $4$ guests at home but I have only two chairs. Say I want to randomly choose two guests to sit and of course rest two stand along with me. ${4 \choose 2}$ gives me number of ways to make two groups. But these groups are ordered. One of the groups sits and the other group stands.
But if I am only interested in finding number of ways I can randomly pair my $4$ guests to play a doubles game of ping pong, there are only $\frac{1}{2!}{4 \choose 2} ~$ ways. By the way, my four guests are Alpha, Bravo, Charlie and Delta. Here are three ways to pair them -
$\{$Alpha, Bravo$\}$,$\{$Charlie, Delta$\}$
$\{$Alpha, Charlie$\}$,$\{$Bravo, Delta$\}$
$\{$Alpha, Delta$\}$,$\{$Bravo, Charlie$\}$
So for the given question to find desired probability, you either need to have number of unordered groups in both the numerator and the denominator or number of ordered groups in both. To make the numerator ordered, note that $A$ and $B$ can swap places to be in named groups and so you multiply ${2n-2 \choose n-1}$ by $2!$.