How many ways can you choose $4$ teams of $2$ from $8$ people. How many ways can you choose $4$ teams of $2$ from $8$ people.
My thoughts were that you have $8$ slots to be filled so you have $8!$ ways to arrange them but this overcounts by a factor of $2$ since a team $a,b$ is the same as a team $b,a$. So in total there would be $$\frac{8!}{2}=20160$$ ways.
Is this right?
 A: We assume the teams are not labelled. Line up the $8$ people in a row, in order of age, or weight, or student number.
The leftmost person has $7$ ways to choose her team mate.
For every such choice, the leftmost person not yet teamed up has $5$ ways to choose her team mate.
And for every such choice, the leftmost person not yet teamed up has $3$ choices, for a total of $(7)(5)(3)$.
A: The first team can be selected in $\binom{8}{2}$ ways, the second team can be selected in $\binom{6}{2}$ ways, and the third team can be formed in $\binom{4}{2}$ ways, and the remaining $2$ form the last team of $4$. Thus there are: $\binom{8}{2}\binom{6}{2}\binom{4}{2}$ ways.
A: $8!$ overcounts more than just $a,b$ and $b,a$, it also accounts for the order of things and in your question order does not matter.  I believe the answer is  ${8\choose2}\cdot{6\choose 2}\cdot{4\choose 2}$?
A: It depends on what you mean by "teams".  If your "teams" are distinguishable then there are 8!/2^4.  If your "teams" are generic indistinguishable groups then you still need to divide by 4!
One way to see this is to imagine 4 empty glasses in a row.  Put the integers 1,2,3,4,5,6,7,8  into a pitcher and stir them up really good.  Pour two integers into each glass.  If the glasses are distinguishable then we just need to divide the 8! ways to stir up the pitcher by 2*2*2*2.  If the glasses are indistinguishable then we still need to divide by 4! 
