So, we saw the following theorem on the partitions of integers:
a) The number of distinct vectors $(n_1,...,n_r)$ of positive integers (i.e $n_1,..,n_r > 0$) satisfying $n_1 + ... + n_r = n $ is $${n-1\choose r-1}$$
b) The number of distinct vectors $(n_1,...,n_r)$ of non-negative integers (i.e $n_1,..,n_r \geq 0$) satisfying $n_1 + ... + n_r = n $ is $${n+r-1\choose n}$$
I tried approaching this theorem using the stars and bars representation.
The way I understand b) is the following. Suppose we have $n$ stars which are lined up, and we want to create $r$ groups by placing separators (thus we need $r-1$ separators). Here, we ask ourselves: "How many ways are there to form $r$ groups?" We can answer this the following way. First we can choose where to place the separators, then place the stars. There is ${n+r-1\choose r-1}$ ways to place the separators and $1$ way to place the stars (since they are indistinguishable). Thus, by the multiplication principle, the number of ways to form $r$ groups is ${n+r-1\choose r-1} \cdot 1$ (which is the same as ${n+r-1\choose n}$, as we could have placed the stars first and then the separators).
My problem is when I try to find a similar explanation for a). The way I see it is that since $n_i > 0 $ for all $i=1,..,r$, this means that I cannot have two bars without any stars in between them (i.e we are forming $r$ groups, but with the condition that each group must contain at least one star). I do not understand how I could count the number of ways I can form $r$ groups (where each group contains at least one star) nor why this number is equal to ${n-1\choose r-1}$
Hint: put one star in each of the $r$ areas separated by the $r-1$ bars, then count how many ways you can place the remainging $n-r$ stars in the $r$ areas.