Number of solutions of $x+y+z=10$ 
The number of different solutions $(x,y,z)$ of the equation $x+y+z=10$ where each of $x, y$ and $z$ is a positive integer is $36$.

How to derive this answer?
I know that $x, y$ and $z$  have to be $1\leq x,y,z\leq8$ and $x,y,z\in\mathbb Z.$ This said me $8+7+6+5+4+3+2+1=36.$
But I can't understand the process of counting the solutions.
 A: Case 1: $x = 1$, $z = 9 - y$, $1 \le y \le 8$.  That's the $8$.
Case 2: $x = 2$, $z = 8 - y$, $1 \le y \le 7$.  That's the $7$.
etc,
A: Let $x = k + 1$, $y = m + 1$, and $z = n + 1$. Then you solve the equation: $k + m + n = 7$, and count the number of non-negative integers solution to this equation. It is a well known answer that it is: $\binom 9 2 = 36$.
A: A way of picturing this (illustrating what Robert Isreal describes) is to consider the tilted plane $ \ x + y + z = 10 \ . $  In the graph below, the lines of intersection with the coordinate planes are marked in red.

The number of positive integer solutions to the equation is the number of integer lattice points that lie on that plane and within the triangle indicated (since we are not to include zeroes in the sums).  On the level $ \ z = 1 \ , $ there would then be eight such lattice points, seven on level $ \ z = 2 \ , $ and so on down to one on level $ \ z = 8 \ . $  This gives the number of permitted solutions as the eighth "triangular number", given as 
$$ \left( \begin{array}{c} 8 + 1 \\ 2 \end{array} \right) \ = \ \frac{9!}{2! \ 7!} \ = \ 36 \ \ .$$
One could generalize this to say that the number of positive integer solutions to $ \ x + y + z \ = \ N \ $ (with integer $ \ N \ge 3 \ $ ) is the $ \ (N-2) th \ $ triangular number, $ \left( \begin{array}{c} N - 1 \\ 2 \end{array} \right) \ . $
