How many ways can you build a football/soccer team? I'm reading a book on the history of soccer. There are several ways to position a team, for example, four defenders, four midfielders and two forwards is the common 4-4-2. How many ways can you build a team, given that there are 10 players that move because the keeper is always in the same position?
I made some schemes:
10
1-9
1-1-8
1-1-1-7
1-1-1-1-6
1-1-1-1-1-5
1-1-1-1-1-1-4
1-1 1-1-1-1-1-3
1-1-1-1-1-1-1-1-2
1-1-1-1-1-1-1-1-1-1

1-2-7
1-2-1-6
1-2-1-1-5
1-2-1-1-1-4
1-2-1-1-1-1-3
1-2-1-1-1-1-1-2
1-21-1-1-1-1-1-1

1-3-6
1-3-1-5

Then I thought of another scheme, one that would group all possible combinations of diagrams from the number of lines that are willing players on the court.
L1 10
L2 1-9,9-1,2-8,8-2,3-7,7-3,6-4,4-6,5-5
L3 

1-1-8,1-8-1,8-1-1,1-2-7,1-7-2, 2-7-1,2-1-7,7-2-1, 7-1-2,1-3-6,1-6-3 .....
I can not find the formula to translate this, if it is correct.
In looking for an answer I found this site. Could you guide me to a book or article to solve my problem?

For variety, here's another approach.
Your initial approach looks like you settled on the following way to enumerate all ways to make a list of positive integers that adds to 10:


*

*Write down all lists that start with 1 followed by a list that adds to 9

*Write down all lists that start with 2 followed by a list that adds to 8

*...


where you repeat the same approach to write down all lists that add to 9, and so forth.
If $f(n)$ is the number of lists that add to $n$, then your approach suggests
$$ f(n) = f(n-1) + f(n-2) + \ldots + f(1) + 1 $$
We can simplify this recurrence by comparing $f(k)$ with $f(k+1)$: they share most of the same terms, so if we subtract them, most of them cancel out:
$$ f(k+1) - f(k) = f(k) $$
and thus
$$ f(k+1) = 2 f(k) $$
This recurrence is very easy to solve: we're just doubling $f(k)$ every time we move to the next value of $k$. Because $f(1) = 1$, we thus conclude $f(n) = 2^{n-1}$.
