combinatorics
This was a question from my textbook. This is how I approached it.
For the first part, you can do c(10,5) so you choose 5 of the 10 to be on a team.
For the second part, you have 10 ways to choose the "best player" and you multiply it by c(10,4) to pick the other 4 players. However, I am not sure how you can exclude the "worst player". Although this sounds subjective, I believe the question is saying that the "best" and "worst" players are distinct, whereas the other players aren't.
Another good way of solving problems like this is to set up a correspondence with a problem you already know how to solve. In this case, imagine that you've got a sorted set of six non-consecutive numbers $a_1, a_2, \dots a_6$ between 1 and 49. What does it look like? Well, the first number $a_1$ is essentially arbitrary; it can't be greater than a certain value (i.e. 39) or else there isn't 'room' for five more non-consecutive numbers above it, but we can ignore that for now. The second number $a_2$ has to be greater than $a_1+1$ — that's what it means to be nonconsecutive, after all — so we know that $a_2-1$ (call this $b_2$) is greater than $a_1$. Similarly, $a_3$ has to be greater than $a_2+1$, so $a_3-1$ is greater than $a_2$, and $a_3-2$ is greater than $a_2-1 = b_2$; we can call this $b_3$. And so on, and so forth, until we define $b_6 = a_6-5$. But this correspondence works both ways — given the $b_n$ it's easy to get the $a_n$ — and so we have a one-to-one correspondence between our non-consecutive sets of numbers and an ordinary combination (but with a different upper bound - can you see what it should be?). It takes a while to build this sort of instinct, but learning how to spot these correspondences is the most basic tool for proving most combinatorial identities.
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):
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
and the weights of the teams
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.
Best Answer
You seem to be making an incorrect assumption about how the problem should be interpreted. The way I read the problem there is a very clear single person who is considered "the best player" and no choice is needing to be made as to who it is.
A rephrasing of the question:
How many $5$-element subsets are there of the set $\{1,2,3,\dots,9,10\}$ such that $1$ is included and $10$ is not included in the subset.
Hint:
Compare this to the problem of asking how many $4$-element subsets there are of the set $\{2,3,4,\dots,9\}$ there are. Do you see what this has to do with your original problem?