<s>[INST] How many 3 -digit numbers can be formed from 0 to 8 whose sum is equal to 20 for repetition and non-repetition? I have a range from $0$ to $8$ . I want to find that how many 3-digit numbers can be formed whose sum is equal to $20$.
Non-Repetition of numbers:
$$
5+7+8=20  \\
$$
Repetition of numbers:
$$ 
4+8+8=20, \\ 
6+7+7=20, \\
8+6+6=20,   
$$
I want to know the number of ways that I can represent 20 using the sum of 3- positive integers. In repetition case, there is only 1 way which is mentioned above but in non-repetition case, we can also do it on paper but I want the general solution.       
 [/INST] The problem can be transformed into a couple of related problems as follows:

Problem 1: Consider the non-repetition problem with no restriction on range. Here we are to find natural numbers $0\leq a<b<c$ such that:
$$a+b+c=20$$
Given appropriate $x,y,z\geq0$ we could reformulate this in the following way:
$$
\begin{align}
a&=x \\
b&=x+1+y\\
c&=x+1+y+1+z
\end{align}
$$
so the equation becomes
$$
3x+2(1+y)+1+z=20 \\
\Updownarrow\\
3x+2y+z=17
$$

Constraint 1: To make use of the results from problem 1, we need to identify the cases where $c\leq 8$. For this to be the case, we must have:
$$
c=x+1+y+1+z\leq 8 \\
\Updownarrow \\
x+y+z \leq 6
$$

Solutions 1: In order to search for solutions to problem 1 subject to constraint 1, we can start from a particular solution:
$$
(x,y,z)=(5,1,0) \implies (a,b,c) = (5,7,8)
$$
and try adding neutral vectors (vectors with $3x+2y+z=0$) such as $(-2,3,0)$ and $(0,-1,2)$ to this in order to produce further solutions. Indeed
$$
(5,1,0)+(-2,3,0)=(3,4,0)\implies(a,b,c)=(3,8,9)
$$
would have been another solution, had it not been for the constraint $x+y+z\leq 6$. Since adding $(-2,3,0)$ or $(0,-1,2)$ increases $x+y+z$ by one and since $(x,y,z)=(5,1,0)$ already has sum $6$, soon one realizes that this is the only solution. Note that $(-2,3,0)$ and $(0,-1,2)$ form a basis for the nullspace of the problem since the nullspace defined by $3x+2y+z=0$ must be two-dimensional.

Problem 2: Consider the repetition case $0\leq a\leq b\leq c$. This can be described via:
$$
\begin{align}
a &= x \\
b &= x+y \\
c &= x+y+z
\end{align}
$$
so we have:
$$
3x+2y+z=20
$$

Constraint 2: Here we have:
$$
c=x+y+z\leq 8
$$

Solutions 2: Again we start from a particular solution:
$$
(x,y,z) = (6,1,0) \implies(a,b,c)=(6,7,7)
$$
Note that $x+y+z=7$ so this is a valid solution. But as soon as we add $(-2,3,0)$ or $(0,-1,2)$ we increase the sum $x+y+z$ by one, so we have very little wiggle room. In fact we can add one of those exactly once, so the only two other solutions must be:
$$
\begin{align}
(6,1,0)+(-2,3,0) &= (4,4,0) \implies &&(a,b,c)=(4,8,8) \\
(6,1,0)+(0,-1,2) &= (6,0,2) \implies &&(a,b,c)=(6,6,8)
\end{align}
$$
I hope this makes sense!
</s>