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!
 A: Converting to $$\alpha'+\beta'+\gamma'+\delta'=16$$ 
The number of ways to distribute $n$ identical things among $r$ different persons/parameters = $\binom{n+r-1}{r-1}$ 
The number of ways of distributing $16$ among these $4$ i.e. $n=16$ and $r=4$, are $\binom{16+4-1}{4-1}=\binom{19}{3}$.
But there are conditions that $$0\le\alpha'\leq 4, 0\leq \beta'\leq 6, 0\leq \gamma'\leq 5, 0\leq\delta'\leq 6$$
So, we need to subtract the number of ways of distribution in which these weren't followed.
For example: To calculate the number of sequences in which $\alpha^{'}\geq 5$, we will first allot $5$ out of $16$ to $\alpha^{'}$ and then follow the same formula as follows (so as to distribute $11$ among $4$)
$$\binom{16-5+4-1}{4-1}=\binom{14}{3}$$
Similarly for $\beta'\geq7$
$$\binom{16-7+4-1}{4-1}=\binom{12}{3}$$
$\gamma'\geq6$
$$\binom{16-6+4-1}{4-1}=\binom{13}{3}$$
and $\delta'\geq7$
$$\binom{16-7+4-1}{4-1}=\binom{12}{3}$$
So, it becomes $$\binom{19}{3}-\binom{14}{3}-\binom{12}{3}-\binom{13}{3}-\binom{12}{3}$$
Finally, the cases in which at least two conditions failed are also to be subtracted but since they are considered already in the cases in which at least one condition failed, we actually need to subtract them once. Why once? Cases of $\alpha^{'}\geq 5$ included cases of any of the other three ($\beta'\leq6;\gamma'\leq5;\delta'\leq6$) not following together with or putting similar argument for other three conditions, we can contemplate that all in all $4\times 3=12$ terms are already subtracted and that we need to now add the "at least two disobeyed condition cases"($^4C_2=6$ in number). 
So, calculating the cases to be added:
For the condition for $\alpha$ and $\beta$: $\binom{16-5-7+3}{3}=\binom{7}{3}$
Similarly for others...
The final answer comes out to be 
$$\binom{19}{3}-\binom{14}{3}-\binom{12}{3}-\binom{13}{3}-\binom{12}{3}+\binom{7}{3}+\binom{6}{3}+\binom{6}{3}+\binom{8}{3}+\binom{7}{3}+\binom{5}{3}$$
This is a widely used approach to questions like this and is pretty common! You may feel like a lot of things working but the format is pretty simple. $\text{Total - at least one condition followed or whatever + at least two conditions... upto which it is not possible}$. (like here it wasn't possible for any three conditions to hold simultaneously obviously)
A: 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
$$
\binom{16+3}{3}-\binom{16-5+3}{3}-\binom{16-6+3}{3}-2\binom{16-7+3}{3}+\binom{16-11+3}{3}+2\binom{16-12+3}{3}+\\2\binom{16-13+3}{3}+\binom{16-14+3}{3}
$$
A: One approach would be to first express it as a problem in positive integers; let $a=\alpha-1$, $b=\beta+2$, $c=\gamma$ and $d=\delta-3$. Then you want to find the number of nonnegative integers $a\leq4$, $b\leq6$, $c\leq 5$, $d\leq6$ such that
$$a+b+c+d=16.$$
Note that, since $a+b+c\leq15$ we have $d\geq1$, and similarly $b\geq1$. So by the substitutions $b'=b-1$ and $d'=d-1$ we in fact want to find the number of nonnegative integers $a\leq4$, $b'\leq5$, $c\leq5$, $d'\leq5$ such that
$$a+b'+c+d'=14.$$
It is not hard to see that for any choice of $a$, $b'$ and $c$ with $a+b'+c\geq9$ there is a unique $d'$ satisfying the above, and there is no such $d'$ otherwise. So it suffices to count the number of nonnegative integers $a\leq4$, $b'\leq5$, $c\leq5$ satisfying $a+b'+c\geq9$. We can count them as follows:
There is one solution to $a+b'=9$, and any value of $c$ yields a solution. This yields a total of $1\times6=6$ solutions.
There are two solutions to $a+b'=8$, and any value of $c\geq1$ yields a solution. This yields a total of $2\times5=10$ solutions.
There are three solutions to $a+b'=7$, and any value of $c\geq2$ yields a solution. This yields a total of $3\times4=12$ solutions.
There are four solutions to $a+b'=6$, and any value of $c\geq3$ yields a solution. This yields a total of $4\times3=12$ solutions.
There are five solutions to $a+b'=5$, and any value of $c\geq4$ yields a solution. This yields a total of $5\times2=10$ solutions.
There are five solutions to $a+b'=4$, and any value of $c\geq5$ yields a solution. This yields a total of $5\times1=5$ solutions.
For $a+b'<4$ there are no solutions. Adding all the above together yields a total of $6+10+12+12+10+5=55$ solutions.
