Number of combination of incremental numbers for a number I had been trying to calculate the number of possible combinations for:
$x_1+x_2+x_3=20$ where $x_i > x_{i-1}$ and $x_i$ are positive whole numbers including zeros.
I had managed to solve a simpler version of this (with only two variables) with star and bar method. But I am having issue with anything with more than two variables.
$x_1 + x_2 = 20$ 
I know that $x_2 \gt x_1$ therefore $x_2 \gt 20 - x_2$ which simplifies to $x_2 - 10 \gt 0$ or $x_2 - 11 \ge 0$
Let $x'_2 = x_2-11$
The new equation will be $x_1 + x'_2 = 9$
This can be solved using star and bar method which gives me 9 combinations.
I was not able to apply the same technique to the equation on top.
It'll be great if someone and hint me in the right direction. I am sorry if I format the equation wrongly. Math is not my expertise.
 A: Start with Stars and Bars to see that there are $\binom {22}2=231$ ways to do it if you ignore the inequality.
Naively, we'd like to divide by $3!=6$ to put them in increasing order, but this won't work because you might have ties between the $x_i$. So let's deal with the ties.
It is impossible for all three to be equal, since $20$ is not divisible by $3$.
If $x_1=x_2$ there are $11$ possible triples, since $x_1$ can be anything from $0$ to $10$ and $x_1$ determines the triple.  Similarly, there are $11$ possible triples with $x_1=x_3$ and another $11$ with $x_2=x_3$.  Thus $33$ "tied" triples all in all.
It follows that there are $231-33=198$ triples with no ties.
Now the naive idea works and we can divide by $6$ to put the remaining triples in order.  thus the answer is $$\boxed {\frac {198}6=33}$$ 
A: You have to compute the so-called number of partitions of n into 3 distinct parts, where 0 is allowed as a part. You may take a look at OEIS-A001399 for more references. There is a nice closed formula:
$$\text{round}( n^2/12 ).$$
Therefore for $n=20$ we get $\text{round}(20^2/12 )=\text{round}(100/3)=33.$
