We can model this problem as a network as follows. We can describe the outcomes for teams of size 2 as a weighted graph, where an edge between two players indicates the number of wins that pair has had. The toy example in the question results in the graph:
In this graph, members A, B and D are equivalent (formally, there are automorphisms of the graph that can map $x$ to $y$ for all $x,y \in \{A,B,D\}$). Thus, there is insufficient information to distinguish whether team $AB$ is better than $AD$ or $BD$. The graph does suggest, however, that any team involving $C$ will be poor.
As for a general method for deciding which pair might make a good team, we will need to make some choice as to what data would indicate a good team. I.e., we will need to decide on some kind of network measure for the edges in the network. A lot of network theory follows along these lines: there can be numerous intuitively good choices of measures, but can give results that contradict one another. Moreover, practical constraints, such as the ability to compute the measures, also plays a large role in what measures to use.
Here's one simple possibility (there's probably much more sophisticated measures in the literature):
- For each player $p$, let the player weight $W(p)$ be the sum of the weights of the edges it is an endpoint of.
- We assign a team $pq$ the team weight $W(pq)=W(p)+W(q)$. The larger the weight, the better.
To illustrate, let's suppose players $A,B,C,D$ paired up and played some more games, and the resulting graph looks like this:
We can calculate the weights of the players
- $W(A)=4$,
- $W(B)=6$,
- $W(C)=1$, and
- $W(D)=3$.
and the weights of the teams
- $W(AB)=4+6=10$,
- $W(AC)=4+1=5$,
- $W(AD)=4+3=7$,
- $W(BC)=6+1=7$,
- $W(BD)=6+3=9$, and
- $W(CD)=1+3=4$.
By this measure, we would conclude that $AB$ is the best team.
One might argue that this measure does not capture some important property of real-world data. This is to be expected of such a basic model. The next step is to develop a better model that incorporates the missing property (which, in turn, will have its own deficiencies). Then we improve that model, and repeat until we're at a point where we're generally satisfied.
The above will extend to $k$-player teams by using $k$-uniform hypergraphs.
You want to combine all possibilities into one big team arrangement. All selections (winger, midflier...) are independant one from another, so you should multiply them : the answer to your problem is ${11 \choose 2}\times{7 \choose 2}\times{5 \choose 2}\times{9 \choose 3}$
If you want to be convinced of it, just look at this small example.
I have 2 choices for my main plate, and 3 choices for my desert. How many different menus are there?
If I pick the first main plate, I have 3 choices for desert.
If I pick the second main plate, I have 3 choices for desert.
For every choice of main plate (and there are 2 of them), I have 3 choices for desert: I actually have $2 \times 3$ choices in total. Hence, multiplication is what you're looking for.
Best Answer
a) Pick who the referee is in $11$ ways. Then among the ten people leftover, one will be the youngest. Pick who the other four members of that player's team is in $\binom{9}{4}$ ways. We get a total of $11\binom{9}{4}=1386$ ways.
b) Pick who the referee is in $11$ ways. Simultaneously pick the two captains in $\binom{10}{2}$ ways. Pick who the other four members of the youngest captain's team is in $\binom{8}{4}$ ways. We get a total of $11\binom{10}{2}\binom{8}{4}=34650$ ways.
Alternate approach:
a) Arrange five a's, five b's, and a c in a line in $\frac{11!}{5!5!1!}$ ways. Have the players go to either team $A$, team $B$, or referee position according to the arrangement of the a's, b's, and c's compared to their names. Now, "forget" which team was actually team $A$ and which team was team $B$ by dividing by two. This gives us $\frac{11!}{5!5!1!}\cdot\frac{1}{2}=1386$ ways.
b) Arrange four a's, four b's, one A, one B, and one C in a line in $\frac{11!}{4!4!1!1!1!}$ ways. The a's will correspond to which players are normal members of team $A$, the A will correspond to the player who is captain for team $A$ etc... Finally, "forget" which team was actually team $A$ and which was team $B$ by dividing by two. This gives us $\frac{11!}{4!4!1!1!1!}\cdot\frac{1}{2}=34650$ ways.
Similarly, one could instead just take the answer to part (a) and then pick who from the youngest person's team is captain and pick who the captain is for the other team, making it so there is $5\cdot5\cdot1386=34650$ ways.