<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] One way is to consider this as a multiset of 3 types of elements, 7 in total (as there is at least 1 of each type, substract that). So the number of solutions is:
$$
\left(\!\!\binom{3}{7} \!\! \right) = \binom{3 + 7 - 1}{7} = \binom{9}{7} = 36
$$
</s>