Simple combination with repetition problem

The "brute force" approach is given in ashleyde's answer: count the solutions to $$ x_1+x_2+x_3=11-x_4\tag{1} $$ for $x_4\in\{0,1,2,\dots,11\}$; that is, $$ \sum_{x_4=0}^{11}\binom{13-x_4}{2}\tag{2} $$ $(2)$ counts all solutions to $$ x_1+x_2+x_3\le11\tag{3} $$ Another way of looking at $(1)$ is to count the solutions to $$ x_1+x_2+x_3+x_4=11\tag{4} $$ which is $$ \binom{14}{3}\tag{5} $$ $(4)$ says that the role of $x_4$ is to enumerate the equations in $(1)$.

This gives a combinatorial proof that $(2)$ and $(5)$ are equal.