How many ways can 2m objects be paired and assigned to m boxes? In how many ways can $2m$ objects be paired and assigned to $m$ boxes? 
In that post the questioner claims that there are $\frac{(2m)!}{2^m}$ to pair $2m$ and assign them to $m$ boxes.
My attempt :-
First we need to find number of ways to pair $2m$ objects. 
That is $\frac{(2m)!}{2! * (2m-2)!}$ = $\frac{2m * (2m - 1)}{2}$ = $m * (2m - 1)$
After pairing there are $m$ objects left, and there are $m!$ to permute them in $m$ boxes
Thus the answer should be $m! * m * (2m -1)$ ways.
Which is wrong. what i am doing incorrectly, please anyone correct me.  
 A: Line up the $m$ boxes in some order from left to right.  We choose two of the $2m$ objects to place in the first box, choose two of the remaining $2m - 2$ boxes in the second box, choose two of the remaining $2m - 4$ boxes in the third box, and so forth until we place the final two objects in the $m$th box.  We can do this in 
\begin{align*}
\binom{2m}{2}&\binom{2m - 2}{2}\binom{2m - 4}{2} \cdots \binom{2}{2}\\
& = \frac{(2m)!}{2!(2m - 2)!} \cdot \frac{(2m - 2)!}{2!(2m - 4)!} \cdot \frac{(2m - 4)!}{2!(2m - 6)!} \cdots \frac{2!}{2!0!}\\
& = \frac{(2m)!}{2^m}
\end{align*}
ways. 
The number $\binom{2m}{2}$ is the number of ways to choose a particular pair of the $2m$ objects to place in a box.  It is not the number of ways to pair the objects.  To do that, we can line up the objects in a row.  There are $2m - 1$ ways we can choose an object to pair with the first object in line.  This leaves us with $2m - 2$ unpaired objects. We have $2m - 3$ ways to choose an object to pair with the first unpaired object remaining in the line.  This leaves us with $2m - 4$ unpaired objects.  We have $2m - 5$ ways to choose an object to pair with the first unpaired object remaining in the line.  Continuing in this way, we obtain
$$(2m - 1)(2m - 3)(2m - 5) \cdots 1$$ 
ways of placing the $2m$ objects in $m$ pairs.  
Multiplying the expression for the number of pairs by $m!$ yields 
\begin{align*}
(2m - 1)&(2m - 3)(2m - 5) \cdots 1 \cdot m!\\ 
& = (2m - 1)(2m - 3)(2m - 5) \cdots 1 \cdot m! \cdot \frac{2^m}{2^m}\\
& = (2m - 1)(2m - 3)(2m - 5) \cdots 1 \cdot \frac{(2m)(2m - 2)(2m - 4) \cdots 2}{2^m}\\
& = \frac{(2m)!}{2^m}
\end{align*}
which is the number of ways of placing the pairs in $m$ boxes.
A: The number of ways that $2m$ objects can be grouped into $m$ boxes where each box contains 2 objects is:
$$ \binom{2m}{2_{b1}, 2_{b2} ...  2_{bm}} = \frac{(2m)!}{2_{b1}!2_{b2}! ... 2_{bm}!} = \frac{(2m)!}{2^m} $$
where $b1$ is the first box, and $bm$ is the last box.
A: The wrong point in your think is that the number of ways to paired $2m$ objects is not $\frac{(2m)!}{2!(m-2)!}$, this number is for chose a pair of $2m$ objects.
In fact, the number of ways to paired $2m$ objects is: 
Put all $2m$ on a row, that is $(2m)!$ ways. Now, each 2 objects, makes a pair, but how the order of the pairs doesn't mater we divided by $m!$ (because we have $m$ pairs). But the order of the objects in each pair doesn't mater, so we divided by 2 in each pair.
The final number is: $\frac{(2m)!}{m!2^m}$.
