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 $C(a,b)$..?
 A: 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 $$\binom{3 + 11 - 1}{11} = \binom{13}{11} = \dfrac{13\cdot 12}{2} = 6\cdot 13 = 78$$
A: imagine 11 balls in a row and two blocks which you will place somewhere. you insert the blocks before, after or between the balls and then you assign values to $x,y,z$ in the following way: $x$ is the number of balls from the beginning of the row up to the first block, $y$ the number of balls between the two blocks and $z$ number of balls from the second block up until the end of the row. you will easily see that the number of ways in which you can place the blocks is equal to the number of different triplets $x,y,z$. Do you know how to compute the number of possible distributions of blocks?
A: Okay let us write a solution to $a+b+c+d+e=10$ a different question, just incase it is homework.
Each solution will have the form:
||||-|---||||| <-> 4As 1B 0Cs, 0Ds, 5Es 

How many different ways can we arrange 10 |s and 4 (4=5-1) -s?
Each arrangement of these |s and -s is a valid solution.
$$\frac{(10+4)!}{4!10!}=\frac{14!}{10!4!}$$
