How many solutions does the equation $x+y+z=11$ have? How many solution does $x+y+z=11$ have where $x, y, z$ are non-negative integers. In light of the restrictions, its clear that $x,y,z \in \{0,1,2,..11\}$. So, at face value I would assign a value for $x$ and determine the different combinations that $y$ and $z$ can hold. For example,
For $x=0$, we have $y+z=11$. With writing them out I found that there are $12$ different assigned combinations for $y$ and $z$ that satisfy the equation. For $x=1$, I got $11$. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is $1+2+3+4+5+6+7..+12=78$. I was wondering if there is an easier method perhaps with combinations equation $C(a,b)$..?

This is a version of the classic stars-and-bars problem in combinatorics.
For any pair of natural numbers $n$ and $k$, the number of distinct $n$-tuples of non-negative integers whose sum is $k$ is given by the binomial coefficient $$\binom{n + k - 1}{k}$$ 
Here, $n = 3$, and $k = 11$, giving you $$\binom{3 + 11 - 1}{11} = \binom{13}{11} = \dfrac{13\cdot 12}{2} = 6\cdot 13 = 78$$
