Question on players/teams and the different combinations If I have $8$ players ABCDEFGH and wish to form two teams. I want to find out how many different combinations are possible with order not mattering.
For example team 1 being ABCD is the same as DCBA but not the same as ABCF so that would be $2$ combinations right there.
I really hope what I'm asking is clear.
 A: Counting the number of combinations of m objects among n objects can be done using the binomial coefficients (you can find plenty of information about that on internet if you want).
In your case, you want to pick 4 objects (people in your case) among 8, so the total number of combinations is going to be $\binom{8}{4} = \frac{8!}{8!(8-4)!}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}=70$.
Now, you might want to divide that amount by 2, because ABCD vs EFGH is the same as EFGH vs ABCD. It depends if you want to count the number of different teams (in that case it is 70), or the number of possible matches (in that case it is 35, for the reason just mentioned).
A: Assuming that You have asked this question and You are not very Familiar with combinatorics , I will answer it this way. You said the order does not matter , If order may have mattered You might have been looking for Permutations But Yes You are right we have to find the possible Combinations for forming a team of two using these 8 players.
A Simple way of doing this is to assume all cases Like this

*

*Out of these 8 players 1 Player is in a Team and the other 7 players which are left are in another Team. So Your Task is to first find ways to Choose this Single Player Team and once you do that The other 7 players will automatically Fit into the other Team.
Mathematically Taking Choosing A Single Player out of 8 is written like this $8 \choose  1$.

*Now It's time to Form a team of 2 and place it in one team and Other 6 Players Left in Another. Mathematically it is this way $ 8\choose 2$.

*$3$ Players in One Team $5$ in Another mathematically=> $8 \choose 3$
... and so on
So it Will go up until You Will form a team like this where you have all the 8 players going to one team and No Players in another written like $ 8 \choose 8$.
So sum of all these combinations is gonna be
$ 8 \choose 0$ + $ 8 \choose 1$ + $ 8 \choose 2$ + $ 8 \choose 3$ .... $ 8 \choose 8$= $2^8$
So this will be your answer thinking it this way.
Another way to think is Like this  For Every Player A , B , C , D , E , F, G , H there are two possibilities for each player they can choose either team to go for. So it can also be written straight away like this 2 possibilities for all 8 players
$2*2*2*2*2*2*2*2$ => $2^8$
