Simple combination with repetition problem How many solutions are there to the inequality
$x_1 + x_2 + x_3 ≤ 11$,
where $x_1, x_2$ and $x_3$ are non-negative integers? [Hint: Introduce
an auxiliary variable $x_4$ such that $x_1 + x_2 + x_3$ +
 $x_4$ = 11.]
Would my reasoning be correct if I let $x_4 = x_1 + x_2 + x_3, x_4 = 11$
Then proceeded as normally with $14\choose11$? I'm a bit unsure of how the auxiliary variable comes to play.
 A: 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.
A: If you use a "matchstick" approach -  using 11 x's and 3 |'s, count the number of x's to the left of each pipe to determine $x_1$, $x_2$, and $x_3$:
|||xxxxxxxxxxx => 0 + 0 + 0 
xxx|||xxxxxxxx => 3 + 0 + 0  
x|x|x|xxxxxxxx => 1 + 1 + 1 
xxxxxxxxxxx||| => 11 + 0 + 0 
In the above examples, the number of x's that are not used in the equation is what $x_4$ is equal to.  So, in the first example, $x_4 = 11$.  In the second example $x_4 = 8$.  Basically, you use the pipe symbols to partition the x's into 4 compartments, the size of each compartment being equal to $x_1$, $x_2$, $x_3$, $x_4$.
One other thing to note, the number of ways to arrange the 3 |'s and 11 x's should be a formula that is familiar to you.
