I want to solve this problem with generating function :
How many solutions(non-negative) possible for the equation ${2x}+{3y}+{7z}={r}$$(r \ge0)$ such that :
1)$x,y,z \ge0$
2)$0 \le z\le 2 \le y \le 8 \le x$
I want to solve this problem with generating function :
How many solutions(non-negative) possible for the equation ${2x}+{3y}+{7z}={r}$$(r \ge0)$ such that :
1)$x,y,z \ge0$
2)$0 \le z\le 2 \le y \le 8 \le x$
$\textbf{Hint: }$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 $z^{16}(1-z^2)^{-1}.$
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^2 + \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) $