<s>[INST] The number of integral solution of $\alpha+\beta+\gamma+\delta$=18 such that.. Question

The number of integral solution of the equation $$\alpha+\beta+\gamma+\delta=18$$, with the conditions:
$1\leq\alpha\leq5$; ${-2}\leq\beta\leq4$; $0\leq\gamma\leq5$ and $3\leq\delta\leq9$ is $k$. Find $k$.

$$Attempt$$
I tried to do to it by giving the minimum number required to the variables to generate new variable to whom then nil can also be given and hence the equation I wrote is $$x+y+z+v=16$$ But, now, I had a problem in adjusting the higher limit for every variable for accounting each of them. Afterwards, I don't know what to do.
               *Any hints or suggestions?*

Thanks for helping me!
 [/INST] Here is a proof using the method of generating functions. By adjusting the constraints we may write the problem as finding the number of integral solutions to 
$$
\alpha'+\beta'+\gamma'+\delta'=16
$$
with $0\le\alpha'\leq 4, 0\leq \beta'\leq 6, 0\leq \gamma'\leq 5, 0\leq\delta'\leq 6$. The number of solutions is then the coefficient in the expansion of 
$$
\begin{align}
F(x)&=(1+x+x^2+x^3+x^4)(1+x+\dotsb+x^6)^2(1+x+\dotsb+x^5)\\
&=\frac{1-x^5}{1-x}\frac{(1-x^7)^2}{(1-x)^2}\frac{1-x^6}{1-x}\\
&=\frac{x^{25} - x^{20} - x^{19} - 2 x^{18} + x^{14} + 2 x^{13} + 2 x^{12} + x^{11} - 2 x^7 - x^6 - x^5 + 1}{(1-x)^4}.
\end{align}
$$ 
For the coeffficient extraction use the fact that
$$
\sum_{n=0}^\infty\binom{n+3}{3}x^n=(1-x)^{-4}
$$
and linearity of coefficient extraction to get that the number of solutions is
$$
\binom{16+3}{3}-\binom{16-5+3}{3}-\binom{16-6+3}{3}-2\binom{16-7+3}{3}+\binom{16-11+3}{3}+2\binom{16-12+3}{3}+\\2\binom{16-13+3}{3}+\binom{16-14+3}{3}
$$
</s>