The number of nonnegative integer triples with sum equal to $12$

Question

How many triples $x,y,z$ satisfy $x+y+z=12$, where $x,y,z\ge 0$ are integers?

Note that these are not ordered: $ (12,0,0)$ and $(0,12,0)$ are treated as same.

Progress

I have tried the formula ${13\choose 2}$ to select values of $x$ and $y$ from $0$ to $12$ as $z$ would keep changing according to $x$ and $y$. (That is, $z=12-x-y$.) The problem is that that certain combinations of $x$ and $y$ would give value of $x+y$ greater than $12$. I can't seem to get around this.

Answer 1

This is a standard stars an bars argument. Present $12$ pictorically as follows.

$\underbrace{************}_{12}$. You want to split this number into three parts so that order matters.

hence you want to add two barriers such as this example, which represents $3+4+5$

$\underbrace{***}_3|\underbrace{****}_4|\underbrace{*****}_5$.

Or this one for $0+0+12$

$||************$.

How many of these arrangements are possible? There are $14$ objects, (12 stars and $2$ bars) hence $14$ positions and we must select the positions the bars will take. There are $\binom{14}{2}$ ways to do this, hence the answer is $\binom{14}{2}=\frac{14!}{(14-2)!2!}=\frac{14\cdot13}{1\cdot2}=7\cdot13=91$ solutions.

Answer 2

From Wikipedia:

The number pk(n) of partitions of n into exactly k parts is equal to the number of partitions of n in which the largest part has size k.

In our case, the number of partitions of n into exactly three parts (where there might be a part equal to zero) is equal to the number of partitions of n in which each part might have a size up to 3.

From "Some Famous Problems of the Theory of Numbers" by Hardy. G. H:

The number of partitions of n in which all parts are 1, 2 or 3 is the nearest integer to $ \dfrac{(n + 3)^2}{12}$

Thus it is equivalent to

The number of partitions of n into at most three parts is the nearest integer to $ \dfrac{(n + 3)^2}{12}$

When you choose a number equal to 0, assume you chose two parts instead of three (or one part if you choose two numbers equal to 0).

Finally for your case the result would be round $\left(\dfrac{(n + 3)^2}{12}\right)$ = round $\left(\dfrac{(12 + 3)^2}{12}\right) = $round$(18.75) = 19$

Extra:

For the same problem but with $x, y, z \geq 1$, we could subtract floor$\left(\dfrac{n}{2} - 1\right)$, since:

the number of partitions of n in which all parts are 1 or 2 (or, equivalently, the number of partitions of n into 1 or 2 parts) is $\lfloor{\frac {n}{2}+1\rfloor}$