Solutions to $a+b+c=12$, $a, b, c \in \mathbb{N}_0$ Let $a, b, c \in \mathbb{N}_0$. If $a+b+c=12$, how many solutions $(a,b,c)$ satisfy the equation? 
Is the answer: 

$729$

 A: Force $a=0$, then, $b+c=12$ has $13$ solutions. Force $a=1$, then, $b+c=11$ has $12$ solutions. In general forcing $a=n$ yields $13-n$ solutions. Since $a$ can run from $0$ to $12$, the total number of solutions is $$\sum_{n=0}^{12}(13-n)=91$$ 
In general the equation $a+b+c=m$ has $\binom{m+2}{2}$ solutions in $\mathbb{N}_0$.
A: Hint: Imagine that you put $12$ coins on a table, side by side on a horizontal line with some spacing between them. You have $2$ pencils each of which you can put between any two of the coins. The $2$ pencils separate these $12$ coins to three groups. These three groups represent the numbers $a$, $b$, and $c$. We now count..
A: Use generating functions.  The generating function for $\mathbb{N}_0$ is 
$$1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x}$$
That is, that the number of ways to write an integer $n$ as an integer is $1$, and the series whose coefficient of $x^n$ is that value is $\frac{1}{1-x}$.
The generating function of the sum of three integers is then that function raised to the third power:
$$\frac{1}{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}{2} x^n.$$
A: If N0 refers to natural numbers {1,2,....}, then the answer is 54 
else if it refers to whole number {0,1,2,...}, then the answer is 87 if I am not wrong.
