<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] As Trismegistos says, this is, using $[z^k]$ to denote the coefficient of $z^k$ in what follows:
$$
[z^{10}] (z + z^2 + \cdots + z^{10})^3
$$
Now use the formula for the sum of a geometric series,
then expand the cube in the numerator and the denominator as a power series. Note that the terms in $z^{10}$ and higher in the first factor have no effect on the result:
\begin{align}
[z^{10}] z^3 \left( \frac{1 - z^{10}}{1 - z} \right)^3
  &= [z^7] (1 - 3 z^{10} + 3 z^{20} - z^{30}) 
         \cdot \sum_{k \ge 0} \binom{-3}{k} (-1)^k z^k \\
  &= [z^7] \sum_{k \ge 0} \binom{-3}{k} (-1)^k z^k \\
  &= (-1)^7 \binom{-3}{7} \\
  &= \binom{7 + 3 - 1}{3 - 1} \\
  &= \binom{9}{2} \\
  &= 36
\end{align}
</s>