In how many ways can $2m$ objects be paired and assigned to $m$ boxes? Been refreshing my statistics knowledge and came across this problem while going through a book.

How many ways can $2m$ objects be paired and assigned to $m$ boxes?

The book claims the answer is
$$ \frac{(2m)!}{2^m} $$
Which makes sense, however, it further claims that if we want to arrange the pairs without assigning them to boxes, since there are $m!$ ways to assign the $m$ pairs to $m$ boxes, the answer is:
$$ \frac{(2m)!}{m!2^m} $$
Can someone explain why we have to divide by $m!$ if we are no longer assigning the pairs to boxes?
 A: We divide by $m!$ because the arrangement (order and which boxes they are put into) matters. We can arrange the boxes in $m!$ ways and we would have a different arrangement. So we have to divide by $m!$ from the previous answer.
A: Let $k$ be the number of ways to pair up the objects.
If we want to pair them up and then assign the pairs to $m$ boxes, we can do this in 2 steps:
1) Pair them up, which can be done in $k$ ways.
2) Assign the $m$ pairs to the boxes, which can be done in $m!$ ways.
Therefore $\displaystyle k(m!)=\frac{(2m)!}{2^m},\;$ so $\displaystyle k=\frac{(2m)!}{(m!)2^m}$
A: If $m = 1$, we have a set $\{1,2\}$ and there's only $1$ unordered pair we can form.
If $m = 2$, we have a set $\{1,2,3,4\}$ and there are $3$ pairings we can form, namely,
$$\{\{1,2\}, \{3,4\}\}$$
$$\{\{1,3\}, \{2,4\}\}$$
$$\{\{1,4\}, \{2,3\}\}$$
For a general $m$, we have a set $\{1,2,\dots,2m\}$ and the number of pairings we can form is the number of partitions of size $2$ of the set, which is given by the following double factorial
$$\begin{array}{rl} (2 m - 1)!! &= (2m-1) \cdot (2m-3) \cdot (2m-5) \cdots 3 \cdot 1\\\\ &= \dfrac{(2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{(2m-2) \cdot (2m-4) \cdot (2m-6) \cdots 4 \cdot 2}\\\\ &= \dfrac{(2m) \cdot (2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{(2m) \cdot(2m-2) \cdot (2m-4) \cdot (2m-6) \cdots 4 \cdot 2}\\\\ &= \dfrac{(2m) \cdot (2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{2^m \cdot m \cdot(m-1) \cdot (m-2) \cdot (m-3) \cdots 2 \cdot 1}\\\\ &= \dfrac{(2m)!}{2^m \cdot m!}\end{array}$$
The total number of assignments of $m$ partitions of size $2$ to $m$ boxes is
$$m! \cdot (2m-1)!! = m! \cdot \dfrac{(2m)!}{2^m \cdot m!} = \dfrac{(2m)!}{2^m}$$
which is also given by
$$\binom{2m}{2} \binom{2m-2}{2} \binom{2m-4}{2} \cdots \binom{2}{2} = \dfrac{(2m)!}{2^m}$$ 
