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?
 A: You can solve it using generating functions. Generating function for this is:
$(x^1 + x^2 + ... + x^{10})^3$. You have to find coefficient of $x^{10}$
A: 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}
A: Hint: Try and prove that the number of $x,y,z\geq 0$ for which $x+y+z \leq 10$ is the same as the number of $x,y,z,w\geq 0$ for which $x+y+z+w = 10$
A: 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}
A: Another possibility is using stars and bars, i.e., you have a line of 10 stars (the sum of the variables), which you have to divide into 3 parts (the individual values) by bars, so that there are no adyacent bars (no variable can be 0). This means distributing $3 - 1 = 2$ bars among the $10 - 1$ spaces between stars, i.e., $\binom{9}{2} = 36$.
A: 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:
$$
\left(\!\!\binom{3}{7} \!\! \right) = \binom{3 + 7 - 1}{7} = \binom{9}{7} = 36
$$
A: Following up on Trismegistos' idea of using generating functions: note that limiting the values to 10 really makes no difference, each variable can at most take the value 8 (two at 1, other one is 8). But if there is no limit, we can just write:
\begin{align}
[z^{10}] (z + z^2 + \cdots)^3
  &= [z^{10}] z^3 (1 + z + z^2 + \cdots)^3 \\
  &= [z^7] (1 - z)^{-3} \\
  &= (-1)^3 \binom{-3}{7} \\
  &= 36
\end{align} 
A: First, we transform the question $x+y+z=10$ for every positive integer to $a+b+c=7$ for every non-negative integer.
We look at a more general question of finding the number of solutions in non-negative integers to the equation $ a + b + c = n $. Since the value of $a$ can be any non-negative integer $0,1,2,3, \ldots, i , \ldots $ (see note below), we can represent this as the generating function $$ A(x) = 1 + x + x^2 + \cdots + x^i + \cdots . $$  We have the same generating function for the possible values of $b$ and $c$.  So our generating function for the number of solutions is $A(x) \times B(x) \times C(x) = [A(x)]^3$. However, finding this product could be extremely tedious.
We instead transform $ A(x) $ into the rational function $ \frac{1}{1-x} $, which we recognize from the sum of a geometric progression. Thus, we are interested in $ [A(x)]^3 = \frac{1}{(1-x)^3} $. This can be expanded using the negative binomial theorem, which gives
$$ \frac{1}{(1-x)^3} = {2 \choose 2} + {3 \choose 2} x + { 4 \choose 2 } x^2 + \cdots + { i+2 \choose 2 } x^ i + \cdots. $$
Therefore, the answer when $n=7 $ is given by $ { 9 \choose 2 } = 36$. This agrees with what we know from the stars and bars method. $_\square$
Note: It might be confusing why we allow $a$ to be any non-negative integer, even those which are larger than $n$, which clearly would not lead to a solution. Consider what would happen if we let $a = n+1$ or $a = n+2$ or any larger integer:  In the final product of polynomials, the exponents of these terms would be larger than $n,$ so they will not contribute to the term we want, which has exponent $n.$
