If $x+y+z=n$ has $(n-1)(n-2)/2$ positive integer solutions, how many does $2x+3y+6z=1200$ have?

Question

From combinatorics I know that it's combinations formula $\binom{n-1}2$ for $x+y+z=n$. But how can I deduce it for something with coefficients, like

$2x+3y+6z=1200$

For context, how similar problem (by books authors opinion) can be solved:

Problem. In how many ways can five identical balls be placed in three different boxes drawers so that no drawer is empty?

Solution. Let's first put one ball in each drawer - then no drawer will be empty. We are left with two balls, which must be placed in the three drawers in an arbitrary way. The number of such arrangements is the number of sequences of two zeros and two ones, i.e. $\binom{2}4$.

We can reason in another way. Let's put five balls in a row. There can only be partitions only in the spaces between the balls. So we need to put two of the dividers in any two of the four spaces, and we can choose two positions out of the four by $\binom{2}4$ in a number of ways.

Task. How many solutions in positive integers does the equation x + y + z = 5 have?

Solution. This is exactly the previous problem, since none of the variables now can't equal zero. The equation has $\binom{2}4$ solutions in the natural numbers. It is not difficult to solve the following problem: (here goes my problem)

See for example here for an example like yours., namely for $4x + 2y + 5z = 100$. See also this post, asking the same as you. — Dietrich Burde, May 04 '24 at 18:22
@DietrichBurde, thanks for mentioning this solution! However, it looks a bit tedious and not in combinatorics way — Michael, May 04 '24 at 18:29
The beginning of the question has almost nothing to do with how to answer the second part, as far as I can see. — Thomas Andrews, May 04 '24 at 18:31
@ThomasAndrews I saw this problem in my combinatorics book, right after the problem "how many ways to put 5 apples in 3 boxes". Authors say that solution to this problem has the same combinations nature — Michael, May 04 '24 at 18:34
@DietrichBurde Well, the way I know how to d9 this is a common technique in combinatorics - generating functions. — Thomas Andrews, May 04 '24 at 18:34
So, tell us the context in the book - what did the chapter just discuss? — Thomas Andrews, May 04 '24 at 18:35
write $x=3t$ and $y=2u$ then divide through by $6$ What is $t+u +z ; ; ? ; ; $ — Will Jagy, May 04 '24 at 18:39
@WillJagy, well it's not 200 :-), and it again brings to, much simplier though, $2x+3y+z=200$ — Michael, May 04 '24 at 18:44
@ThomasAndrews Have a look at my answer, the two are intimately related. (replying to this comment) — Mike Earnest, May 04 '24 at 18:46
@Michael no, $t+u+z = 200$ is how to continue. There is now an answer writing this out, he calls the new variables $x', y'$ — Will Jagy, May 04 '24 at 18:47

score 1 · Accepted Answer · answered May 04 '24 at 18:45

The general strategy is to look at the equation modulo $2$ and modulo $3$, in order to restrict the possibilities for some the variables, and thus simplify the problem.

Writing $$ 2x=1200-6z-3y, $$ we see that the RHS is a multiple of $3$, so the LHS must be a multiple of $3$ as well. But $2$ and $3$ are coprime, so we have proven $x$ must be a multiple of $3$. That is, we now know $x=3x'$ for some integer $x'$.

Similarly, looking at $$ 3y=1200-2x-6z, $$ we conclude that $y$ is a multiple of $2$, so $y=2y'$ for some integer $y'$.

Now, substituting $3x'$ for $x$ and $2y'$ for $y$ in the original equation, we get $$ 6x'+6y'+6z=1200. $$ You can now count the number of solutions to this equation, using the same method as counting solutions to $a+b+c=n$.

If $x+y+z=n$ has $(n-1)(n-2)/2$ positive integer solutions, how many does $2x+3y+6z=1200$ have?

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