1. ## Combinatronics Problem...

"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. If the pairing is done at random what is the probability that there are no offensive-defensive pairings?"

My first thoughts were that there are (40Cr2)=720 possible pairings although i'm told its much larger than this...

Any ideas? Any prods in the right direction would be much appreciated.

2. Originally Posted by sebjory
"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. If the pairing is done at random what is the probability that there are no offensive-defensive pairings?"
The are $\displaystyle \frac{40!}{(2^{20})(20!)}$ random pairings.
Can you finish?

3. using patterns i would imagine the amount of OOs and DDs is

(20!)/((2^10))(10!)) and that the probability would be this divided by your number.

However could you explain how you arrived at this expression?

Many thanks

(p.s. is there a tutorial anywhere on how to use the Math Input function on this forum... the thing denoted by the capital Sigma)

4. Here is a very quick and dirty way to explain this.
Say we have a class of 20 freshmen.
If we divide them up into four study groups: biology, mathematics, literature, and history.

That can be done in $\displaystyle \binom{20}{5}\binom{15}{5} \binom{10}{5} \binom{5}{5} =\frac{20!}{(5!)^4}$ ways. ($\displaystyle \binom{N}{k}$ N choose k.)
Those are known as ordered partitions. Because it makes a difference to a student which group he/she is in.

Think a class of 20 freshmen is calculus.
If we divide them up into four groups to review a just returned test.
Those are known as unordered partitions because only content matters.
This can be done $\displaystyle \frac{20!}{(5!)^4(4!)}$ ways.
The details are a bit much to into.
But I hope you can begin to get the idea.

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