The question in the book is as follows:
A football team consist s of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. If the pairing is done at random, what is the probability that there are no offensive–defensive roommate pairs?
My Solution was:
Sample Space = $40 \choose 2,2,....,2 $ , there are 20 2's.
Now, we pair the offensive players together, and do the same for the defensive players.
Event = $20 \choose 2,2,...,2$$\times$ $20 \choose 2,2,...,2$, there are 10 2's each for both the mutichoose's.
Now probability = $\frac{Event}{sample space}$
However the authors calculated the unordered pairs, which goes as follows:
Sample Space = $40 \choose 2,2,....,2 $ $\times$ $20!$
Event = $20 \choose 2,2,...,2$ $\times 10!$$\times$ $20 \choose 2,2,...,2$$\times 10!$
My questions is: Why is choosing ordered pairs a mistake in this case?