How many ordered quadruples $(a,b,c,d)$ satisfy $a+2b+3c+4d=420$ where $a,b,c,d$ natural numbers or $\mathbb{N}^+$ How many ordered quadruples $(a,b,c,d)$ satisfy $a+2b+3c+4d=420$ where $a,b,c,d$ natural numbers or $\mathbb{N}^+$
I know we need to find the coefficient of $x^{420}$ in
$$
\frac{1}{(1-x)} \cdot \frac{1}{(1-x^2)} \cdot \frac{1}{(1-x^3)} \cdot \frac{1}{(1-x^4)}
$$
Could someone help me to finish this?

Besides the fact that I don't know anything about generating functions, one thing that I dislike about that approach is that you simply seem to be transferring the computational difficulty.
That is, I have seen people set up an answer, based on generating functions, and then use computer software to calculate the pertinent coefficient.  It seems to me that you could just as readily have the computer manually count all such solutions.
The following approach suffers from the same difficulty.  That is, I can set up the computations, but then there won't be any elegant way of finding the final answer.  My understanding is that absent generating functions, the approach given below is standard.


How many ordered quadruples $(a,b,c,d)$ satisfy $a+2b+3c+4d=420$ where $a,b,c,d$ natural numbers or $\mathbb{N}^+$

I am assuming that $a,b,c,d$ must each be a positive integer.
For any real number $r$, let $\lfloor r\rfloor$ (i.e. the floor function) denote the largest integer $\leq r.$
Work backwards.
Suppose that you have the equation 
$a + 2b = n,$ where $n$ is a positive integer, 
and you want to know how many solutions there are in positive integers $a,b$.
Here, $b$ can take on any value from $1$ through 
$\displaystyle \left\lfloor \frac{n-1}{2}\right\rfloor.$
So, let $f_2(n)$ denote 
$\displaystyle \left\lfloor \frac{n-1}{2}\right\rfloor ~: ~n \in \Bbb{Z^+}.$

Now, consider the equation 
$a + 2b + 3c = n.$ 
Here, $c$ can take on any value from $1$ through 
$\left\lfloor \frac{n-1}{3}\right\rfloor.$ 
For each such value for $c$, there will be $f_2(n-3c)$ solutions. 
So, the number of solutions to 
$a + 2b + 3c = n$ 
is
$$f_3(n) = \sum_{c=1}^{\left\lfloor \frac{n-1}{3}\right\rfloor} f_2(n - 3c).$$
In a very similar fashion, the final computation will be
$$\sum_{d=1}^{\left\lfloor \frac{420-1}{4}\right\rfloor} f_3(420 - 4d). \tag1 $$

The difficulty with the answer given in (1) above, is so what.  Again, all that has been done is to organize the very cumbersome calculations.  You still wouldn't want to try calculating this manually.
