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Homework Statement
A football team consists of 20 offensive and 20 defensive players to be randomly paired as roommates. What is the probability there are no offensive-defensive player pairs?
Homework Equations
r-groups of n-objects outcomes = [itex] \frac{n!}{n_1!n_2! \dots n_r!}[/itex]
The Attempt at a Solution
This is actually worked out as an example in my book (A First Course in Probability, Ross), but there is something I don't understand. It says the number of ordered pairs is [itex]\frac{40!}{(2!)^{20}}[/itex] and the number of un-ordered pairs is [itex]\frac{40!}{20!(2!)^{20}}[/itex]. and the latter is used in the denominator of the probability calculation. I don't understand the context of ordered vs un-ordered and why the 2nd is used in the denominator.
It's sort of like saying there are [itex]\frac{52!}{(13!)^4}[/itex] ordered sets of bridge hands and [itex]\frac{52!}{4!(13!)^4}[/itex] un-ordered sets of bridge hands. What does that mean?