Positive solutions

Yes, it is a bars and stars problem, and the number of solutions:

$N = \left( \begin{matrix} 12-1 \\ 3-1 \end{matrix} \right) = \left( \begin{matrix} 11 \\ 2 \end{matrix} \right) = \dfrac {11\times 10} {2} = \boxed{55}$

Since balls and urns requires for x, y, and z, to be non-negative, and the problem is asking for positive integers, adjustment is necessary. From each x, y, and z, subtract 1 resulting in subtracting 3 from 12 (equals 9). Then use the balls and urns technique where x + y + z = 9. (9+2) C 2 = 55.

To begin with, $x$ , $y$ , and $z$ are not equal to 0 and are mot negative (given). As a result, $x$ , $y$ , and $z$ must at least be equal to 1. Thus, we would take away 3 of the 12 stars (9 stars). Note that we also have 2 bars to separate $x$ , $y$ , and $z$ , so add that to the current 9 objects. Now, we have 11 objects to rearrange, and the way to organize $x$ , $y$ , and $z$ is by where the bars occupy within the 11 object slots. Thus, we would calculate ${11} \choose {2}$ , which is $\boxed{55}$ .

By stars and bars, the number of non-negative integral solutions will be $\binom{n+k-1}{n} = \binom{n+k-1}{k-1}$ which, in this case, should equal $\binom{3+12-1}{2} = \binom{14}{2}=91$ . However, we have also included the possibilities with $a$ and/or $b$ and/or $c$ being equal to $0$ , which they cannot be because $0$ isn't a positive integer. We must subtract those combinations.

Case 1: Exactly one of $a,b,c$ equals $0$ . In this case we will have to subtract the number of solutions to the equation $x_1+x_2=12$ , $3$ times, because there are $3$ ways to choose a pair: there is $a,b;a,c;b,c$ . By stars and bars, this is $3\binom{13}{1}=13*3$ . We have, again, counted $0$ , so we must subtract another $6$ solutions from this. There are, therefore, $3\binom{13}{1}-6$ combinations here.

Case 2: Exactly two of $a,b,c$ equal $0$ . It can be seen that we must subtract another three combinations here.

Our grand total is therefore $91-(3\times13-6)-3$ , which works out to $55$ .

Use combination technique and use formula n-1Cr-1 where n is 12 and r is 3