# How many solutions possible for the equation ${2x}+{3y}+{7z}={r}$

I want to solve this problem with generating function :

How many solutions(non-negative) possible for the equation $${2x}+{3y}+{7z}={r}$$$(r \ge0)$ such that :

1)$x,y,z \ge0$

2)$0 \le z\le 2 \le y \le 8 \le x$

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If $r$ is given (and $x,y,z$ are unknowns that comprise the solution), you should explain that in the question. The answer "how many solutions" then depends on $r$, right? –  hardmath Dec 27 '12 at 17:22
Also, you explicitly state that the variables are supposed to take integer values, otherwise the question becomes quite boring... –  Ben Millwood Dec 27 '12 at 17:29
I edited question. –  Martin Dec 27 '12 at 17:41

$\textbf{Hint: }$Note that the number of soloutions is given by the coefficient of $z^r$ in the expansion of the following expression : $$\left((z^2)^8+(z^2)^9+(z^2)^{10}+\cdots \right)\left((z^3)^2+(z^3)^3+\cdots +(z^3)^8\right)\left((z^7)^0+(z^7)^1+(z^7)^2\right)$$ and the first expression can be written as $z^{16}(1-z^2)^{-1}.$
Thanks. In part 1, we have :$$\left((z^2)^0+(z^2)^1+(z^2)^{2}+\cdots \right)\left((z^3)^0+(z^3)^1+\cdots \right)\left((z^7)^0+(z^7)^1+(z^7)^2+\cdots\right)$$ , How to get $z^{16}(1-z^2)^{-1}.$,,,, and in part 2 , How to get closed form? –  Martin Dec 27 '12 at 17:57
Let $\xi = x^2$, then by the restriction placed on $x$, we have the generating function $$(\xi^8 + \xi^9 + \xi^{10} + \dots) = \xi^8(1+ \xi^2 + \xi^3 + \dots) = x^{16}/(1-x^2).$$ To handle the $y$ term, let $\eta = x^3$, then \begin{align*} (\eta^ + \eta^3 + \dots + \eta^8) &= \eta^2(1 + \eta + \eta^2 + \dots + \eta^6)\\ &= \eta^2 \frac{1-\eta^7}{1-\eta}\\ &= x^6 \frac{1-x^{21}}{1-x^3} \end{align*} Lastly, to handle the $z$ term, let $\zeta = x^3$, then $$(\zeta^0 + \zeta^1 + \zeta^2) = (1-x^21)/(1-x^7)$$