Find all positive integers $x, y, z$,such that $$x+2y+3z=12$$
Using the formula (got in almost similar thread) I got ${k−1 \choose n−1}=55$ solutions are available from the equation where $k=12$ and $n=3$
After reading some similar examples from similar threads I tried to solve it using $\text{Trial and error }$ method-
$12=1.2.2.3$ $$\frac{x+2y+3z}{12}=m$$ $$\therefore 1+2.1+3.3=12 \\1+2.4+3.1=12\\ 2+2.2+2.3=12 \\ 3+2.3+3.1=12 \\ 4+2.1+3.2=12 \\ 5+2.2+3.1=12 \\ 7+2.1+3.1=12$$.
So, out of $55$ solutions I can figure out only 7 solutions.
What is the reason for this difference? If I am wrong anywhere please guide me to a right way. Any help is appreciated.