Q:

Question regarding dividing group of people into ordered pair

This is a very basic counting problem, however I couldn't recall my memory to understand the answer correctly.
From "A First Course in Probability by Sheldon Ross",

Example
  A football team consists of 20 offensive and 20 defensive players. The players are paired in groups of 2 for the purpose of determining roommates....
  There are $\dfrac{40!}{(2!)^{20}}$ ways of dividing 40 players into 20 "ordered pairs" of two each.

I tried a smaller set, says:

Offensive = ${a_1, a_2}$
Defensive = ${b_1, b_2}$

so all ordered pairs are :
$$a_1a_2, a_2a_1, b_1b_2, b_2b_1$$
$$a_1b_1, b_1a_1, a_1b_2, b_2a_1$$
$$a_2b_1, b_1a_2, a_2b_2, b_2a_2$$
So there are 12 of pairs. Now if I use the formula was given above, I got:
$$\dfrac{4!}{(2!)^2} = \dfrac{4.3.2}{4} = 6$$
which is clearly incorrect.
So my question is, should the formula given from the Example be $\dfrac{40!}{{2!}^{19}}$? 

A:

I think you are miss understanding the thing, since it is said:
"There are $\frac{40!}{(2!)^{20}}$ ways of dividing 40 players into 20 "ordered pairs" of two each."
which it means that the ordening of the pairs in the set of pairs of a particular combination of pairs matters that pairs are ordered has not sense since the factorial would not have to be in that case with $2$), i.e. in your case would be the possibilities
$$(a_1,a_2),(b_1,b_2)$$
$$(b_1,b_2),(a_1,a_2)$$
$$(a_1,b_2),(b_1,a_2)$$
$$(b_1,a_2),(a_1,b_2)$$
$$(a_1,b_1),(a_2,b_2)$$
$$(a_2,b_2),(a_1,b_1)$$
If the ordening of the pairs in the set of pairs mattered, but not the ordenind of the particular pairs, then the formula would be
$$\frac{40!}{20!}$$
where the denominator represents the way of ordening the $20$ pairs.
I hope this helps.

