Lottery - how many possibilities to combinate some sum given some draws

Question

I was reading this question on how many possibilities there are to formulate the sum of some number say $k\in \mathbb{N}$ and I was intrigued and tried my way on solving it and got an answer but my answering is rather less mathematical than what I saw on some answers to such a question.
I started with $3$ draws out of $n\in \mathbb{N}$ total numbers.
The lowest number is: $$1 + 1 + 1 = 3$$ So we only have $1$ combination. Adding one we get $3+1=4$ : $$ 1+1+2=1+2+1=2+1+1=4$$ A total of $3$ combinations. Another addition gives us $4+1=5$: $$1+2+2=2+2+1=2+1+2=3+1+1=1+1+3=1+3+1=5 $$ A total of $6$ combinations. Just once more we will add one, so $5+1=6$: $$ 1+2+3=3+2+1=2+3+1=4+1+1=1+1+4=1+4+1=2+2+2=6$$ Leaving us at $7$ combinations. Now one can guess a pattern:
The lowest number at $3$ draws is $3$. Adding $1$ gives you the possibility to add the $1$ to any of the $3$ draws and you will end up with $1\cdot 3=3$ combinations.
Once more you have $3$ combinations of the previous number, here for example of $4$ and add $1$ to each draw, so you have $3$ combinations for the number $5$ to begin with. What you can also do it to split the addition so that you have two $2$'s instead of only $1$. These will again give you $3$ combinations.
Now the only thing is, if you have a number that is a multiple of the starting number, say $c\cdot l, \ c \in \mathbb{Z}_+$, the same as above can be applied, but splitting it will give you the draw of $c\cdot l$ in each draw, i.e. in our example of $6=2+2+2$ only adding one total combination. So my solution of the problem is the following:

Given $\{1,2,...,n\}$ numbers and drawing $l \in \mathbb{N}$. Let $c \in \mathbb{Z}_+$. The possibility to combinate a number $k\in \mathbb{N},\ l \leq k \leq n\cdot l$ is given by: $$\begin{cases}(k-l)\cdot l, \quad k-l\neq c\cdot l\\ (k-1-l)\cdot l + 1, \quad k-l=c\cdot l\end{cases}$$ Now to prove such a statement an induction over $k\in \mathbb{N}$ comes into my mind and so I did and argued almost the same way as above.
The question now is, is my solution correct and if so, is arguing the way I did fine or rather not clear enough? If not, could one be more precise as I am in the induction step for example?

Answer 1

Your problem is quite different from that linked one.

1. you allow repetitions
2. you consider permutations of a sequence as different combinations.
3. you impose no restriction on the values of the terms, except for $x_i\ge1$.

That problem is much simpler and has a well-known closed form solution (see stars and bars): $$ \binom{s-1}{n-1}, $$ where $s$ is the value of the sum, and $n$ is the number of terms (in your example $n=3$).