$a+b+c = 2001$ and $a>b>c>0$ how many ordered triples are possible? $a+b+c = 2001$ and $a>b>c>0$ how many ordered triples are possible? 
I tried substituting these equations
$c+x=b$ 
$b+y=a$
thus,  the equation would be
$3c+2x+y = 2001$
but I do not know how to solve this, is there a more general stars and bars formula?
 A: Since $a,b,c$ are positive integer, they are at least one so we can operate the substitution:
$$ (A+1)+(B+1)+(C+1)=2001 \Rightarrow  A+B+C=1998 $$
Since A,B,C now can be also 0,let's create a bijection between the ordered triples $(a,b,c)$ and a group of binary numbers constructed by this way:
$$ \underbrace{11...11}_{A\text{ times}}0\underbrace{11...11}_{B\text{ times}}0\underbrace{11...11}_{C\text{ times}} $$
Practically $0$ is a variable separator and the $1$ indicates the "units" collocated in each variable. Since the bijection is done, the searched number is equal to the number of permutation of two $0$ and $A+B+C=1998$ ones:
$$N=\frac{2000!}{1998!2!}=1999000$$
Now we must deal with the limitations. Let's find the number $E$ of triples that have 2 or more equal elements. We indicate with $E_{x,y}$ the number of triples such that $x=y$. Then for the inclusion-exclusion principle:
$$E=E_{a,b}+E_{a,c}+E_{b,c}-2E_{a,b,c}$$
Clearly, because of the symmetry of the equation:
$$E_{a,b}=E_{a,c}=E_{b,c}$$
We can easily calculate their value, indeed if two numbers are equal they can vary from 1 to 1000 so:
$$E_{a,b}=E_{a,c}=E_{b,c}=1000$$
Trivially $E_{a,b,c}=1$ because it corresponds to the triple $(667,667,667)$ , so:
$$E=2998$$
These couples clearly don't satisfy the restrictions. So we remain with this number of triples:
$$N'=N-E=1996002 $$
The remaining triples have distinct entries, so they have $3!=6$ permutations of which only one is increasing as requested so:
$$X=\frac{1996002}{6}=332667$$
