The Number Of Integer Solutions Of Equations
An approach is to find the number of distinct non-negative integer-valued vectors $(x_1,x_2,...,x_r)$ such that $$x_1 + x_2 + ... + x_r = n$$
Firstly, considering the number of positive integer-valued solutions.
An approach to solving this problem for positive integer-valued solutions is to imagine that you have $n$ indistinguishable objects lined up and that you want to divide them into $r$ nonempty groups. To do so, you can select $r-1$ of the $n-1$ spaces between adjacent objects as the dividing points. See the diagram below for a visual representation.
$$n\,\, objects\,\,0$$ $$Choose\,\,r-1\,\,of\,\,the\,\,spaces\,\,_\wedge.$$
For instance if you have $n=8$ and $r=3$ and you choose the 2 divisors so as to obtain $$000|000|00$$ then the resulting vector is $x_1 = 3. x_2 = 3, x_3 = 2.$ As there are $n-1\choose r-1$ possible selections, you have the following proposition.
Proposition 1: There are $n-1\choose r-1$ distinct positive integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$x_1 + x_2 + ... + x_r = n, \,\,\, x_i > 0,\,\,\, i = 1,...,r$$
Finally, from Proposition 1 you can obtain the following proposition
Proposition 2: There are $n+r-1\choose r-1$ distinct non-negative integer-valued vectors $(x_1, x_2,...,x_r)$ satisfying the equation $$ x_1 + x_2 + ... + x_r = n$$
Question: I understand all the steps taken prior to Proposition 2, so what I want to know is how is Proposition 2 derived from Proposition 1? I have drawn multiple diagrams using the spaces between objects analogy by adding $r$ extra possible positions for a divider, but none of them hold for all possible vectors.