Why is there a $4$ with $d$ ? Foolish Number !

The problem can be solved with stars=and-bars method with a twice, as $d$ has a $4$ as coefficient.

There are $6$ acceptable values for $d = 0,1,2,3,4,5$ .

When $d = 0$ , then we need to find the number of solutions for $a+b+c = 20$ . This is equivalent to finding the number of solutions for $x+y+z = 23$ using the stars-and-bars method, which requires $x, y, z \ge 1$ , while $a, b, c$ can be $0$ . The number of solutions is given by: $N_0 = \left( \begin{matrix} 23-1 \\ 3-1 \end{matrix} \right) = \left( \begin{matrix} 22 \\ 2 \end{matrix} \right)$ . The number of solutions when $d = n, n = 0,1,2,3,4,5$ is therefore, $N_n = \left( \begin{matrix} 22-4n \\ 2 \end{matrix} \right)$ . Therefore, the total number of solutions:

$N = N_0+N_1+N_2+N_3+N_4+N_5$

$\quad = \left( \begin{matrix} 22 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 18 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 14 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 10 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 6 \\ 2 \end{matrix} \right) + \left( \begin{matrix} 2 \\ 2 \end{matrix} \right)$

$\quad = 231 + 153 + 91 + 45 + 15 + 1 = \boxed {536}$

There is an especially useful function in Mathematica for this kind of problem: (FrobeniusSolve[{a, b, c}, s] for finding the list of all solutions to the equation $a x + b y + c z == s$ , where $a,b,c$ are given positive integers and d is an integer, while $x,y,z$ are non-negative integers to be found.

The formula is FrobeniusSolve[{1, 1, 1, 4}, 20] // Length

Here's a python script to achieve the same:

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def factors_set():
    factors_set = ((i,j,k,l) for i in range(21) for j in range(21) for k in range(21) for l in range(6))
    for factor in factors_set:
        yield factor

def memoize(f):
    results = {}
    def helper(x):
        if x not in results:
            results[x] = f(x)
        return results[x]
    return helper

@memoize        
def linear_combination(n):
    weigh = (1,1,1,4)
    count = 0
    for factors in factors_set():
        sum = 0
        for i in range(len(factors)):
            sum += factors[i]*weigh[i]
        if sum == n:
            count +=1
    print (count)

linear_combination(20)

The equation we have is $20=a+b+c+4d.$ Now,because there is a $4d$ in that equation it makes our job easier as now we know the range of $d$ which is from $0$ to $5.$ $Case\ 1:\\ d=0$ $\Longrightarrow\ a+b+c=20$ .Let us first consider the non-zero cases.We have a theorem for that which is $\dbinom{n-1}{k-1}$ where $n=20(a+b+c)$ and $k=3(a,b,c)$ the proof of this is very simple if one uses stars and bars.Now,let us consider the cases which have one or two or all of $a,b,c=0.$ $Case\ 1:a=0,b=1,2...10,c=19,18...10.$ Now,it is easy to do the permutations. $Case\ 2:a=b=0,c=20.$ This gives a total of $\dbinom{19}{2}$ (non-zero cases+ $60$ (zero cases). $Case\ 2:\\d=1$ $\Longrightarrow\ a+b+c=16.$ Again,let us first consider the non-zero cases $=\dbinom{16-1}{3-1},$ Now let us consider the zero cases $a=0,b=1,2,3....7,c=15,14,13..8.$ We can easily do the permutations giving us $40$ cases.So,total= $\dbinom{15}{2}+48.$ We do the same thing with $d=3$ and see that their is a pattern $\dbinom{19}{2}+60,\dbinom{15}{2}+48,\dbinom{11}{2}+36...$ Last case is $d=5$ which has only one solution $=(a,b,c,d)=(0,0,0,20).$