How many solutions are there to the inequality $x_1 + x_2 + x_3 \le 11$ where $x_1,x_2,x_3$ are nonnegative integers? Hint: introduce a variable $x_4$ such that $x_1 + x_2 +
x_3 + x_4 = 11$.
Solution: $x_4$ is nonnegative, same as count the number of nonnegative solutions to the
equality. It is $\binom{4+11-1}{11} = \binom{14}{3} = 364$
Why did we introduce an extra variable $x_4$? While if we have ($x_1 + x_2 + x_3 = 11$ ) we do not?

First, observe that there are now more solutions to the inequality. When we add the dummy variable, we know introduce a degree of freedom to help account for this new solutions.
Let's try and understand the equation $x_1 + x_2 + x_3 = 11$.This tells us that we have to equal 11, but when we introduce the inequality, we know have to account for $x_1 + x_2 + x_3 = 10$, $x_1 + x_2 + x_3 = 9$, all the way until ... $x_1 + x_2 + x_3 = 0$. How do we account for these? Well, we can introduce another variable that helps us "move" our original fixed value of 11.
Let's define $x_4$ to be greater than or equal to 0. How does this help us? If the first three equal 11, then the fourth will equal 0, if the first three equal 10, then the fourth will equal 1. and so on...
