Probability/Combinatorics Example Question Hi I'm doing an example problem from Sheldon Ross and am confused.  I got the first part of the question so I only listed the 2nd part:
Question:  A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. What is the probability that there are 2i offensive–defensive roommate pairs, i = 1,2,...,10?
20 Offensive (O), 20 Defensive (D) ==> 40 people total and 20 pairs
The question asks for the probability of 2i OD pairs.  If we have 2i OD pairs, then we have $(20-2i)$ OO and DD pairs.  So I think the solution will be of form:  
$\frac{(\mbox{total OD pair combinations})  \cdot (\mbox{total OO Comb)(total DD comb})}{\mbox{total unordered pair combinations}  }$
Total unordered pair combinations = $\frac{1}{20!}\binom{40}{2,...,2}=\frac{40!}{20!(2!)^{20}}$

Total OD pair combinations = $\binom{20}{2i}\binom{20}{2i}(2i!)$ 
This is because out of 20 O's and 20 D's, we pick 2i from each to get 2i OD pairs.  My first question is why do we multiply by $[(2i)!]$ ?  Doesn't this give us a repeated count?  I mean, assume we have 20 O's and 20 D's, then $O_1D_3$ pair is equivalent to $D_3O_1$.  Aren't we over counting by multiplying by $[(2i)!]$??  Because our denomenator is the unordered pair of combinations it seems to conflict to me...
The next part is where I really get confused.  Now we calculate combinations of $(20-2i)$ pairs. So I assume we have to use the binomial theorem again:
$\binom{(20-2i)!}{???}^2$  Apparently, the ???= $(10-i)!$.
Can someone please explain this step by step?  Am I correct to say that $(20-2i)$ = [the number of OO or DD pairs remaining?]  Maybe I'm confused on what i means?  If they say we have 2i OD pairs that means we have 2i O's and 2i D's right?  I can't put these pieces together.
Soln:  $\cfrac{\binom{20}{2i}^2(2i)! \left[ \cfrac{(20-2i)!}{2^{10-i}(10-i)!} \right]^2}  {\cfrac{40!}{2^{20}20! }}   i=0,1,...,10\tag{$\diamondsuit$}  $
I don't know how to adjust latex size so I apologize if it comes out small.
Thanks in advance!
 A: To count the number of possible ways to make $2i$ DO-pairs, we need to count the number of ways to choose the $2i$ offensive players (each of whom will be paired with a defensive player), the number of ways to choose $2i$ defensive players, and the number of ways to match two such selections up in a one-to-one fashion. There are $(2i)!$ ways to accomplish this last. But why? We may as well match them up as follows:

There are $2i$ ways to pair off the offensive player (in the group of $2i$) with the smallest jersey number, there are $2i-1$ ways left to pair off the offensive player with the second-smallest jersey number, and so on, until there is only one choice left for the offensive player with the highest jersey number.

(We won't be able to match up the rest of the players in the same way. Why does it work here and not there?) Hence, there are $$\binom{20}{2i}^2(2i)!\tag{$\clubsuit$}$$ different collections of $2i$ DO-pairs.
Now, we need to see how many ways we can pair off the $20-2i$ remaining offensive players, and how many ways we can pair off the remaining defensive players (these numbers will of course be equal). The number of ordered pair combinations of $20-2i$ players is $$\frac{(20-2i)!}{(2!)^{10-i}}=\frac{(20-2i)!}{2^{10-i}}=\frac{2^i(20-2i)!}{2^{10}},$$ and the number of unordered pair combinations is $$\frac{2^i(20-2i)!}{2^{10}(10-i)!}.$$ The number of ways to pair off all the remaining offensive players with each other and all the remaining defensive players with each other is therefore $$\left(\frac{2^i(20-2i)!}{2^{10}(10-i)!}\right)^2.\tag{$\heartsuit$}$$
A: Imagine that there are $20$ rooms, labelled Room 1, Room 2, and so on. We can pick the occupants of these rooms in $\frac{40!}{2^{20}}$ equally likely ways. This number will be our denominator. 
If there are $2i$ OD pairs, then there are $10-i$ OO pairs and $10-i$ DD pairs. 
We can pick the rooms that will hold the OD pairs in $\binom{20}{2i}$ ways. For each such way, we can pick the rooms that will hold  the OO pairs in $\binom{20-2i}{10-i}$ ways.   By simplifying,  we find that the number of ways to classify the rooms is $\frac{20!}{(2i)!(10-i)!(10-i)!}$.
Now we fill the rooms. The first OD room can be filled in $20^2$ ways. For each such way, the next OD room can be filled in $19^2$ ways, and so on, for a total of $(20)^2(19)^2\cdots (20-i+1)^2$ ways. We may prefer to write this as $\frac{(20)^2}{((20-i)!)^2}$.
Now we fill the $10-i$ OO rooms. The number of ways to do this is $\frac{(20-2i)!}{2^{10-i}}$. And for each of these, there are $\frac{(20-2i)!}{2^{10-i}}$ ways to fill the DD rooms. 
It follows that the number of favourables is 
$$\frac{20!}{(2i)!(10-i)!(10-i)!}\frac{(20!)^2}{((20-i)!)^2}\frac{((20-2i)!)^2}{2^{20-2i}}.$$
