# How many combinations can there be if you have $n$ slots where each value can be $[0,n]$ and the sum of all slots must be equal to $n$?

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: $${2n-1 \choose n-1}$$.