<s>[INST] Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$ 
Now we have
$x + y + z = 10$   with   $x, y, z \leq 10$    (where $x,y,z$ are positive integers and can be the same)
What are the different methods by which we can solve this?
 [/INST] After choosing x and y, z can only have 1 value:
\begin{align}
\sum_{x = 1}^{8} \sum_{y = 1} ^ {9 - x} 1
  &= \sum_{x = 1}^{8} (9 - x) \\
  &= 8 \cdot 9 - \frac {8(8+1)}2 \\
  &= 36
\end{align}
</s>