I have a problem that goes like this, we have $n$ slots that can contain a value $r=[0,n]$ where $n \in \mathbb N$. In addition, the slots must add up to $n$, that is $n=\sum\limits_{i=1}^{n}r_i$. What are the total number of combinations that can be used to satisfy these requirements.
My specific case is where $n=6$. My first reaction was to see what the upper-bound of combinations where the value was allowed to repeat. This would be $$n^n \Rightarrow 6^6 = 46656$$ That would ideally allow me to check and make sure my solution wasn't off base. The next angle I took was to apply the fundamental principal of counting to determine the total number of acts that could be taken in each slot and to multiply them together. I came up with: $$\text{Total # of combinations} = 6(6-r_1)(6-\sum\limits_{i=1}^{2}r_i)...(6-\sum\limits_{i=1}^{n}r_i)\text{ ,where n=5}$$
My logic was that the number of choices of what I can put in the $i^{th}$ slot depends on what I put in the $i-1$ slot. Thinking through the equation I came up with a bit more, I realized that it fell flat on its face. If $r_1=6$ then the total combinations would equal 0 which is clearly not true. I tried looking at the $n=2$ and $n=3$ cases but did not have much luck.
Any tips or general advice would be greatly appreciated!
Note: It has been a while since I have worked on a math problem so please do correct/improve my question.
The answer is the number of solutions to: $$ r_1+r_2+\cdots + r_n = n \text{ where } 0 \leq r_i \leq n $$ To solve it, there is a great method called stars-and-bars. See if you can get the reasoning of that method and try to solve it. The final answer is: ${2n-1 \choose n-1}$.