How many solutions for an equation with simple restrictions
One way of looking at it is this: The number of ways is just the coefficient of $x^{20}$ in the expansion of $$(1+x+x^2+\cdots+x^7)(1+x+x^2+\cdots+x^5)(1+x+x^2+\cdots+x^{20})^{2}$$
Here, each factor represents a term in your equation (here I have written them in the same order). The powers of $x$ in the factor are the values the corresponding variable can take. I limited the last two factors to a maximum power of $20$ as all the terms are non negative.
Now, the logic behind this is that when you multiply two powers of $x$, the powers add up. The coefficient of $x^{20}$ gives the number of ways you could have multiplied various powers from each factor to get a total power of $20$. Notice that each combination of powers from each factor contributes to an increase of one in the coefficient and also represents a unique solution to your equation. Thus, the coefficient of $x^{20}$ in the binomial expansion gives the desired answer.
Each factor can be simplified by using the formulae for the sum of terms in a geometric progression. Then, the $(1-x)^{-n}$ terms can be expanded using the binomial expansion, after which the coefficient can be found easily.
$$(1+x+x^2+\cdots+x^7)(1+x+x^2+\cdots+x^5)(1+x+x^2+\cdots+x^{20})^{2}=\frac{(1-x^8)(1-x^6)(1-x^{21})^2}{(1-x)^4}$$
So, you need to find the coefficient of $x^{20}$ in the expansion of $$(1-x^8)(1-x^6)(1-x)^{-4}$$ The $(1-x^{21})^2$ term can be ignored, as terms with coefficient higher than $20$ are not of any concern. $$(1-x^8)(1-x^6)(1-x)^{-4}=(1-x^6-x^8+x^{14})(1+4x+10x^2+20x^3+\cdots+84x^{6}+\cdots+455x^{12}+\cdots+680x^{14}+\cdots+1771x^{20}+\cdots+\binom{n+3}{n}x^n+\cdots)$$ The coefficient is thus: $$1771-455-680+84=720$$
To determine the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 20 \tag{1}$$ in the non-negative integers subject to the restrictions $x_1 < 7$ and $x_2 < 6$, we subtract the number of solutions in which $x_1 > 7$ or $x_2 > 5$ from the number of solutions of the equation.
A particular solution of equation 1 corresponds to the placement of three addition signs in a row of $20$ ones. For instance, $$1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 1 1 1$$ corresponds to the solution $x_1 = 4$, $x_2 = 5$, $x_3 = 6$, and $x_4 = 5$, while $$+ 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 + 1 1 1 1$$ corresponds to the solution $x_1 = 0$, $x_2 = 9$, $x_3 = 7$, and $x_4 = 4$. Thus, the number of solutions of equation 1 is the number of ways three addition signs can be inserted into a row of $20$ ones, which is $$\binom{20 + 3}{3}$$ since we must choose which three of the $23$ symbols ($20$ ones and $3$ addition signs) will be addition signs.
Suppose $x_1 > 7$. Then $y_1 = x_1 - 8$ is a non-negative integer. Substituting $y_1 + 8$ for $x_1$ in equation 1 yields \begin{align*} y_1 + 8 + x_2 + x_3 + x_4 & = 20\\ y_1 + x_2 + x_3 + x_4 & = 12 \tag{2} \end{align*} Equation 2 is an equation in the non-negative integers with
$$\binom{12 + 3}{3} = \binom{15}{3}$$
solutions.
Suppose $x_2 > 5$. Then $y_2 = x_2 - 6$ is a non-negative integer. Substituting $y_2 + 6$ for $x_2$ in equation 1 yields \begin{align*} x_1 + y_2 + 6 + x_3 + x_4 & = 20\\ x_1 + y_2 + x_3 + x_4 & = 14 \tag{3} \end{align*} Equation 3 is an equation in the non-negative integers with
$$\binom{14 + 3}{3} = \binom{17}{3}$$
solutions.
If we subtract the number of solutions of equation 2 and equation 3 from the number of solutions of equation 1, we will have subtracted those solutions in which $x_1 > 7$ and $x_2 > 5$ twice. Thus, we must add the number of solutions in which $x_1 > 7$ and $x_2 > 5$.
Suppose $x_1 > 7$ and $x_2 > 5$. Then $y_1 = x_1 - 8$ and $y_2 = x_2 - 6$ are non-negative integers. Substituting $y_1 + 8$ for $x_1$ and $y_2 + 6$ for $x_2$ in equation 1 yields \begin{align*} y_1 + 8 + y_2 + 6 + x_3 + x_4 & = 20\\ y_1 + y_2 + x_3 + x_4 & = 6 \tag{4} \end{align*} Equation 4 is an equation in the non-negative integers with
$$\binom{6 + 3}{3} = \binom{9}{3}$$
solutions.
By the Inclusion-Exclusion Principle, the number of solutions of equation 1 in the non-negative integers subject to the restrictions that $x_1 < 7$ and $x_2 < 6$ is
$$\binom{23}{3} - \binom{15}{3} - \binom{17}{3} + \binom{9}{3}$$
The classic way to do this is count the solutions without conditions, then use inclusion-exclusion to deal the cases which violate.
So if $A$ is the set of all non-negative solutions to $x_1+x_2+x_3+x_4=20$, and $A_1$ is the set of such solutions with $x_1\geq 8$ and $A_2$ is the set of solutions with $x_2\geq 6$ then the set you seeking to count is $A\setminus(A_1\cup A_2)$ and:
$$\left|A\setminus(A_1\cup A_2)\right|=|A|-|A_1|-|A_2|+|A_1\cap A_2|$$
By inclusion-exclusion. Now, each of these terms is much easier to compute. For example, $A_1\cap A_2$ is the set of solutions to $x_1+x_2+x_3+x_4=20$ with $x_1\geq 8,x_2\geq 6$, which is equal to the set of non-negative solutions to $y_1+y_2+x_3+x_4=20-14=6$.