How many fours can be formed from eight sports teams? Suppose we have eight teams, let them be A, B, C, D, E, F, G and H. Two teams can form a pair, e.g. AB (BA is exactly the same pair). Then, pairs can make up into fours. For instance, AB, CD, EF and GH is an example of a four. Again, the order of pairs in the four does not matter.
The question is: how many different fours can be formed?
At school, we were solving problems like "In how many ways can you arrange 8 different objects (no identical ones)?" It's 8! = 40320. Here, however, you may arrange/combine them in any way you wish. In the described above situation with sports teams, you cannot use any letter more than once. E.g. BC and CD cannot coexist in the same four.
Basically, the answer is not that important for me; I just wonder which formula should I use or where I can read some information or maybe watch any Youtube videos about combinatorics problems where certain objects don't 'like' the other ones (combinations have restrictions).
 A: Simply denote each pair of team a match
I think the answer is $105$. First, there are $ \binom {8}{2}$ ways to select the first match, then from the remaining $6$ teams, there are $ \binom {6}{2}$ ways to select the second match and similarly $ \binom {4}{2}$ ways to select the third match. Accordingly, the last match is determined.
But each way we draw the four matches is repeated $ 4! =24$ times, thus the answer is:
$ \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} :24 = 105$
A: You are asking for the number of fours, not pairs.
So imagine that there are two numbered courts , you can send the players into the two courts in $\dfrac{8!}{4!4!} = 70$ ways
But since we aren't interested in which court number a player goes to, the required answer is $\dfrac{70}{2!} = 35$ ways

Answer modified according to OP's clarification
OP has clarified that by "fours" is meant four pairs including all $8$ teams.
Then, by the same logic of first imagining $4$ courts, division can be into $\dfrac{8!}{2!2!2!2!} \div4! = 105$ fours
A: There are $7$ possible choices for $A$'s partner. Now let $X$ be the next unpaired player (i.e. $X$ is $B$ unless $B$ was paired with $A$, in which case $X$ is $C$). There are $5$ choices for $X$'s partner. Now let $Y$ be the next unpaired player. There are $3$ choices for $Y$'s partner.
So the answer is $7\times 5\times 3=105$.
A: There are $\binom 84$ ways to choose the $4$ teams that make up your four.  Once you've selected $4$ teams, there are $3$ distinct ways to pair them up within the four.  Thus, the answer is $210=3 \times \binom 84$.
