The problem here is in choosing $n$ objects from $2n$ objects and then arranging them with the remaining objects in such a way to form unique pairs. i.e.,
${2n \choose n}$
This counts the number of ways of selecting $n$ objects from $2n$ objects. The remaining $n$ can be arranged in $n!$ ways.
So, often are we confused that to think what does this ${2n \choose n}n!$ counts.
Imagine, we have $6$ objects. $a,b,c,d,e,f$ we take out $a,b,c$.
We can pair up $a$ with $d,e,f$ i,e $a$can be paired with $3$ $objects$ and $b$ with $2$ and $c$ with $1$ $object$.
which by counting gives $3!$
Therefore, from this we know that each element we choose from ${2n \choose n}$ has $n!$ ways of forming a pair.
But here is the glitch, inside ${2n \choose n}$ we have already counted for all the possible $n$ objects that can be chosen from $2n$ objects.
Therefore ${2n \choose n}n!$, counts for both
$Group$ $ 1$ ${(A,d),(B,e),(C,f)}$ and
$Group$ $2$ ${(d,A),(e,B),(f,C)}$.
Here, the first elements of the subsets of 1st and the 2nd Group have been counted by ${2n \choose n}$ and the second elements of each the group's subsets is just the mere arrangement of $n!$
Now, as we can see clearly the $FIRST$ $AND$ $THE$ $SECOND$ $GROUPS$ are $Equivalent$ but $NOT$ $UNIQUE$.
So, to remove this overcounting of $2$ GROUPS we divide by $2!$
And therefore for each of the combination we have $2!$ which multiplies up to $(2!)^n$ for each of the $n$ elements of the subsets. Which leads to Unique pairs from $2n$ $to$ $n$ $pairs$ is given by $\frac{{2n \choose n}n!}{(2!)^n}$