How many ways to arrange these gifts? (Inclusion-exclusion\derangement) Each one of 30 people has bought 2 identical presents for the poor (every person's gifts are different from everyone else's). All the gifts were put in a large bag. 
In turns, 30 poor people approached the bag and took 2 presents each (the 2 presents were taken simultaneously, not one by one).
In how many ways can they draw gifts so that not one of them had pulled 2 identical gifts?
I was thinking I should take all the ways for withdrawal (60! for the permutations I think) and apply the Inclusion-Exclusion where $Ai$ = poor guy #i pulled 2 identical gifts, $Ai∩Aj$ would be 2 people that drawn identical gifts, etc. But I didn't know how to calculate it.. 
This question is a part of a Combinatorics 101 practice page, under Inclusion-Exclusion Principle \ Derangements
Thank you.
 A: This  problem   presents  no  surprises   and  we  can   use  ordinary
inclusion-exclusion of  the set inclusion  poset. Suppose we  seek the
number of  configurations where  at least $q$  people have  received a
pair of identical gifts.
First we must choose these $q$ people, which gives
$${n\choose q}.$$
View the recipients  as a sequence ordered from  left to right. Second
we must chooose the pairs of identical gifts they receive, which gives
$${n\choose q} \times q!.$$
Finally  distribute the  remaining  gifts  any way  we  like into  the
left-over slots i.e. recipients, getting
$${2n-2q\choose 2,\ldots,2}.$$
Putting it all together we get the inclusion-exclusion formula
$$S_n = \sum_{q=0}^n {n\choose q}^2 
(-1)^q q! \frac{(2n-2q)!}{2^{n-q}}.$$
Starting at $n=1$ we get the sequence
$$0, 4, 48, 1440, 65280, 4348800, 398200320, 48007895040,\ldots$$
Now looking  at the value  for $n=2$ we  see that with this  count the
gifts a  person receives  are ordered as  opposed to being  sets. Note
however that we have the count  $S_n$ where it is already assured that
the gifts  everybody received are  different. Therefore the  orbits of
these configurations when ordered pairs  are turned into sets all have
the same size namely $2^n.$ Therefore with the two gifts being sets we
get the formula
$$T_n = \frac{1}{2^n} \sum_{q=0}^n {n\choose q}^2 
(-1)^q q! \frac{(2n-2q)!}{2^{n-q}}.$$
Starting at $n = 1$ we obtain the modified sequence
$$0, 1, 6, 90, 2040, 67950, 3110940, 187530840, 14398171200,\ldots$$
 Addendum. There is another  approach when we are interested in
the model where  the gifts received form sets  with two elements which
is to replace  the multinomial coefficient with an  application of the
Polya Enumeration Theorem (PET). In particular we need the cycle index
$Z(G_m)$ of  the permutation  group $G_m$ acting  on $2m$  slots where
adjacent slots form sets of two  elements, e.g. $[1,2]$ is the same as
$[2,1].$ This cycle index is
$$Z(G) = \frac{1}{2^m} \sum_{k=0}^m 
{m\choose k} a_2^k a_1^{2m-2k}.$$
Introducing the quantity
$$F_m = [C_1^2 C_2^2\cdots C_m^2] Z(G_m)(C_1+C_2+\cdots+C_m)$$
we thus obtain
$$T_n = \sum_{q=0}^n {n\choose q}^2 
(-1)^q q! F_{n-q}.$$
Starting at $n = 1$ once more we obtain the modified sequence
$$0, 1, 6, 90, 2040, 67950, 3110940, 187530840, 14398171200,\ldots$$
