Phicar’s answer gives you a nice, short calculation if you know about Stirling numbers of the second kind. If not, you can still organize your argument a bit more efficiently.
Suppose that the set $A\cup B$ has $n$ elements; clearly $n$ must be $2,3,4$, or $5$. For each of these four possible values of $n$ we can argue as follows.
There are $\binom5n$ ways to choose the set $A\cup B$. $A$ can be any non-empty proper subset of $A\cup B$. $A\cup B$ has $2^n$ subsets, but one is empty and one is all of $A\cup B$, so there are only $2^n-2$ choices available for $A$. Thus, there are $\binom5n(2^n-2)$ ordered pairs $\langle A,B\rangle$ with $|A\cup B|=n$.
The answer, therefore, is
$$\sum_{n=2}^5\binom5n(2^n-2)=\sum_{n=2}^5\binom5n2^n-2\sum_{n=2}^5\binom5n\;.$$
Now notice that
$$\sum_{n=2}^5\binom5n=\sum_{n=0}^5\binom5n-\binom51-\binom50=2^5-5-1=26\;,$$
and
$$\begin{align*}
\sum_{n=2}^5\binom5n2^n&=\sum_{n=0}^5\binom5n2^n-\binom512-\binom50\\\\
&=\sum_{n=0}^5\binom5n2^n1^{5-n}-10-1\\\\
&=(2+1)^5-11\\\\
&=232\;,
\end{align*}$$
so the final answer is $232-52=180$.
This calculation suggests another elementary way to perform the calculation. If we temporarily allow $A$ and $B$ to be empty, we are in effect counting the ways to split $X$ into $3$ pieces, any of which may be empty. For each of the $5$ elements of $X$ we can put that element into $A$, into $B$, or into $X\setminus(A\cup B)$. This is a $3$-way choice made $5$ times, so there are $3^5=243$ ways to make it. However, some of these splits leave $A$ or $B$ or both empty. We can use an inclusion-exclusion argument to take care of them.
How many of these splits leave $A$ empty? They are the splits that put every element into $B$ or $X\setminus(A\cup B)$, so there are $2^5$ of them. There are also $2^5$ splits leaving $B$ empty, so we have to subtract $2\cdot2^5=64$. However, that subtracts the one split with $A=B=\varnothing$ twice, so we have to add it back in. The final result is
$$243-64+1=180\;.$$
Oh boy, there are a lot more than you can probably imagine.
We want to pick a total of 120 groups. The first group, we must pick 3 elements out of the 360 that are there. There are $360\choose3$ ways to pick them, where $${n\choose k}=\frac{n!}{(n-k)!k!}$$
Next, we would pick the next group with $357\choose3$ ways, etc etc all the way down to $3\choose3$ ways.
The answer is the product of all of these terms, i.e., $$\prod_{n=1}^{120}{3n\choose3}$$Now, we can collapse some terms, since $k=3$ in each combination, and the numerator of each successive term cancels the denominator of the previous. So, the product equals $$\frac{360!}{120!\cdot(3!)^{120}}$$which has 671 digits. I'll post the pastebin for the whole number here. So no, you don't really want a list. You'd overload the entire earth's storage.
As a final note, if you have $n$ items and $k$ objects you want per group, (obviously $n$ is a multiple of $k$), the formula for the number of ways to group the elements is $$\frac{n!}{\frac nk!\cdot(k!)^{n/k}}$$
Best Answer
We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is
$$\dfrac{(2n)!}{2^nn!}.$$
Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$.