Breaking walls

Relevant wiki: Stars and Bars

$\text{the number solution of } x_1 + x_2 + x_3 \leq 20 \text{ minus the number solution of } x_1 + x_2 + x_3 \leq 15\\ \text{So, the answer is } C(20, 3) - C(15, 3) = 685$

$\text {To Solve this Problem , i will be using the help of generating functions } \\ \text{The Max. Value }x_1,x_2,x_3,\text{ Can achieve For the condition to be true is 18 and the min is 1} \\ \text{We need to find the sum of coefficients of }x^{16},x^{17},x^{18},x^{19}\text{ and } x^{20} \text { in : } \\ (x^1+x^2+x^3+\cdots+x^{18})^3 \\ \implies \dfrac{(x^1-x^{19})^3}{(1-x)^3} \implies x^3\dfrac{(1-x^{18})^3}{(1-x)^3}\\ \text{We need to find the sum of coefficients of }x^{13},x^{14},x^{15},x^{16}\text{ and } x^{17} \text { in : }\\ \implies \dfrac{(1-x^{18})^3}{(1-x)^3}\\ \text{We know that : }\dfrac{1}{1-x} = \displaystyle{\sum_{n=0}^{\infty}x^n}\\ \text{Take 1st Derivative : }\dfrac{1}{(1-x)^2} = \displaystyle{\sum_{n=0}^{\infty}nx^{n-1}} \\ \text{Take 2nd Derivative : }\dfrac{2}{(1-x)^3} = \displaystyle{\sum_{n=0}^{\infty}n(n-1)x^{n-2}}\\ \implies \dfrac{1}{(1-x)^3} = \displaystyle{\sum_{n=0}^{\infty}\dfrac{n(n-1)}{2}x^{n-2}}\\(1-.....\text{We don't need the rest of the terms})(\displaystyle{\sum_{n=0}^{\infty}}\dfrac{n(n-1)}{2}x^{n-2}) \\ \implies \text{Required value of }n=15,16,17,18,19 \\ \implies 15\times7+8\times{15}+17\times8+9\times{17}+19\times9 \\ \text{Ans = }\color{#3D99F6}{\boxed{685}}$

Or you could just calculate (15 2)+(16 2)+(17 2)+(18 2)+(19 2)=685, where (x y) is a binomal coeficient example (15 2)=15×14/2