To divide or not to divide: Reducing repeat cases in combinatorics Question:

Find the number of natural number solution of the equation: $$x+y+z+4w=37$$

My attempt:

I'll use: $^{n+r-1}C_{r-1}$
We have, effectively, $x+y+z+4w=30$
Since, $4$ "boxes/lots" are same/alike/identical, and $n=30,r=7$
$$\therefore \text {Answer} =\frac{^{36}C_6}{4!}$$

Video Solution:

$1\leq w\leq8$, we'll make different cases and use: $^{n-1}C_{r-1}$
If $w=1$; $\quad x+y+z=33$; $\quad^{33-1}C_{3-1}$ $=$ $^{32}C_{2}$
If $w=2$; $\quad x+y+z=29$; $\quad^{29-1}C_{3-1}$ $=$ $^{28}C_{2}$
$\dots$
If $w=8$; $\quad x+y+z=5$; $\quad^{5-1}C_{3-1}$ $=$ $^4C_{2}$
$$\therefore\text{Answer=} ^{32}C_{2}{+}^{28}C_{2}{+}^{24}C_{2}+\dots{+}^4C_{2}$$

My doubt:


*

*Why my way of doing is wrong?

*How do I know that when not to do what I did? [Because when I not do it, then I over-work/take fool's way instead of smart way. And when I do it then such cases as this happen]


Please help.
 A: *

*Note that there are differing definitions of natural numbers (see link). The book has taken it to be positive integers, let us go with that


*The stars and bars formula then simplifies to $\Large\binom{n-1}{k-1}$


*To come to your main difficulty, in the equation $x+y+z+4w=37$, we can't take the last term to comprise of four separate boxes, because for every increase of $1$ for $w$, the total increases by $4$. So if we are to use stars and bars, we shall have to do it case by case, starting upwards with $w=1$ , other variables $\geq 1$
https://mathworld.wolfram.com/NaturalNumber.html#:~:text=The%20term%20%22natural%20number%22%20refers,Bourbaki%201968%2C%20Halmos%201974).
A: Note: It's been quite some time since posting the question and OP found the following comments by the user: lulu very helpful. So, they have been compiled here as an answer.

This is hard to follow. You can't treat the $4w$ box as $4$ separate boxes for a number of reasons (not least that the contents of those $4$ boxes might not sum to a multiple of $4$).
As a quick way to see the problem, note that the number of solutions to $4a=7$ is $0$ but there are many solutions to $a_1+a_2+a_3+a_4=7$.
If my original equation was $4a=7$ then that equation would have no solutions, since $7$ is not a multiple of $4$. However, your method clearly yields a non-zero number, as there are solutions to $a_1+a_2+a_3+a_4=7$, and non-zero numbers are still non-zero even after you divide by $4!$
But, if you want an example closer to the one you gave, use $a+4b=7$ which actually has solutions over the natural numbers. We easily see that $b<2$ which means that the only possible case is $b=1$, which leads to the unique solution $(a,b)=(3,1)$. Your method, however, yields $14!×(7+44)=13.75$ which isn't even an integer.
To be correct, your method would need to insist that the four virtual boxes each contain the same value. That would be a true equivalent to the original problem, but of course it is no easier to solve than the original.
After all, which of your cases would, say, $w=2$ correspond to? Would it be $(2,2,2,2)$? Or $(5,1,1,1)$? Or $(4,2,1,1)$? Or $(3,3,1,1)$?
As a general point: if you are given counting problem $A$ and wish to solve it by changing it into counting problem $B$, then you must explain (in detail) the connection between the two problems. Maybe there's a bijection between the solutions, maybe one problem is somehow the complement of the other. But the connection is the key and it must be explained. Here, I see no useful connection between the two problems.
