The Number Of Integer Solutions Of Equations 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.


$$0_\wedge0_\wedge0_\wedge..._\wedge0_\wedge0$$

$$n\,\, objects\,\,0$$
$$Choose\,\,r-1\,\,of\,\,the\,\,spaces\,\,_\wedge.$$

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 $$x_1 + x_2 + ... + x_r = n, \,\,\, x_i &gt; 0,\,\,\, i = 1,...,r$$
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.
For concreteness, let us work with the specific numbers $8$ and $3$ mentioned in the post, though the argument is general.

We have $8$ identical candies, and we want to distribute them among $3$ kids, with some kid(s) possibly getting no candy. Call this Task A.

Task B goes as follows. Distribute $8+3$ candies among the kids, with each kid getting at least $1$ candy. Then take away a candy from each kid.  

It is clear that there are just as many ways to carry out Task B as there are to carry out Task A. And by the analysis of Proposition 1, there are $\binom{8+3-1}{3-1}$ ways to carry out Task B. 

Another way: Imagine $8+3-1$ slots in a row, like this:
$$\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square\quad\square$$
We will put place $8$ candies ("stars") in these slots, with the other $2$ slots serving as separators ("bars"), possibly adjacent. The number of ways of placing the candies is $\binom{8+3-1}{8}$. It is the same as the number of ways of choosing the $2$ slots to be left blank. 

So the number of solutions is $\binom{10}{8}$, or equivalently $\binom{10}{2}$.

In general, the same analysis shows that the number of ways to distribute $n$ candies among $r$ kids is $\binom{n+r-1}{n}$, or equivalently $\binom{n+r-1}{r-1}$.
Well, this particular problem can be solved by thinking in following way:

Step 1: First find the total number of solutions in which none of the variables from x1, x2, and x3 is zero.

Step 2: Then find the number of solutions in which only one from x1, x2, and x3 is zero.

Step 3: Then find the number of solutions in which only two from x1, x2, and x3 is zero.

Step 4: Finally, add all the number of solutions found in the abovementioned steps. This is the final answer.

This method can be generalized and we will arrive at the same formula.

PS: I don't have any idea how to write math equations here, and therefore my contribution may rightly look ugly to some of the contributors here.