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Eight bridge players assembled.They wish to break up in four teams of two each and then pair up the teams to form the two bridge playing groups. In how many ways these grouping can be done?

My Attempt
If we assign the numbers $1 ,2 ,3 .. 8$ to the players then to form a a pair we have to count ordered pair $(a,b)$ . This can be done in $7+6+..1=28$ ways. But But selecting two groups from this $28$ pairs is tricky , as in this pairs single player occurs multiple times , hence once a pair containing two particular players are chosen , then all the other pairs which contain them are out. Here I am stuck up how to proceed further.
Thanks in advance for any help.

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    Is a bridge playing group to be just left as a group, or is it to be further broken up into partners ? – 2017-01-06

1 Answers 1

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The first pair can be chosen in $\binom 82=28$ ways. The second (chosen the first) in $\binom 62=15$ ways. The third in $6$ ways, and there is no choice (or, more precisely, there is one choice) for the last. But if we had chosen the pairs in any other order, we would have the same teams. So the number of ways to make the pairs is $$\frac{28\cdot15\cdot 6}{4!}=105$$

Now when the pairs are made, we have to make pairs with the pairs. By a similar reasoning, we conclude that there are $$\frac{\binom 42}{2!}=3$$ ways to do this.

Then the grouping can be done in $105\cdot 3=315$ ways.

In general, if we have a set with $mn$ elements, there are $$\frac1{n!}\prod_{k=1}^n\binom{km}m$$ ways to make $n$ groups of $n$ elements each.

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    @rugi: You have made a "rugimentary" error in multiplication ! :) – 2017-01-06
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    It appears that there s another typo in the last line "ways to make $n$ groups of $n$ elements each." – 2017-01-06