It turns out there is a simple rephrasing of the problem that makes it easy to count. Rather than write down a list of numbers like (3,5,2) that add to 10, you can write down 10 objects, and place dividers that separate the 10 objects into groups:
* * *|* * * * *|* *
Exercise: convince yourself of the following:
- given any list of numbers that add to 10, you can draw a corresponding diagram like the above
- given any diagram like the above, you can find a corresponding list of numbers that add to 10
- If you have a list of numbers, convert it to a diagram then back to a list, you get your original list
- If you have a diagram, convert it to a list, then back to a diagram, you get the original diagram
There is an easy way to count how many diagrams can be drawn: there are 9 places where we can put a divider, and for each place, we have a choice to put a divider there or not. Therefore, there are $2^9 = 512$ possible choices.
This sort of diagram is called "stars and bars", and there are a number of kinds of problems that can be solved by finding a way to convert back and forth between the problem you want to solve and diagrams like the one above.
If you want to count specifically the number of lists that have a given number of entries, we can do something similar. e.g. if we want there to be exactly three positions (each with at least one person), then we want to count the number of ways to insert two dividers into a list of 10 objects.
Again, convince yourself that you can translate back and forth between lists and diagrams. It can be tricky to get these arguments exactly right sometimes.
If we want there to be $k$ positions, then we want to place $k-1$ dividers into the 9 possible places, and thus there are $\binom{9}{k-1}$ choices.
If you haven't seen binomial coefficients (a.k.a. combinations) before, this is equal to
$$ \binom{n}{r} = \frac{n!}{r! (n-r)!} $$
where $s!$ means the factorial of $s$: that is,
$$ s! = 1 \cdot 2 \cdot \ldots \cdot s $$
(and $0! = 1$)
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
Of the 20 people, first pick 3 of them to be defenders. There are $20 \choose 3$ such ways three people could be chosen. Now there are only 17 people left to pick from that could be a center player. So, there are $17 \choose 5$ ways to pick the next 5 players as centers. Now there are only 12 people left to pick from to be an attacker. So, there are $12 \choose 2$ ways to pick the next 2 players to be attackers.
You have to perform each of these three tasks - pick defenders, then pick centers, then pick attackers. So for each of the $20 \choose 3$ ways to finish the first task, there are another $17 \choose 5$ ways to complete the second task. And for each of the many ways to finish the first and second tasks, there are yet another $12 \choose 2$ ways to finish the third task.
The total is ${20 \choose 3} \times {17 \choose 5} \times {12 \choose 2}$