Number of ordered triples $(a,b,c)$ such that $a+b+c=2019$ **How many ordered triples $(a, b, c)$ are there satisfy the following criteria **

*

*$a,b,c$ are positive numbers

*$a+b+c=2019$

*a, b, c form increasing arithmetic progression

*a , b and c are suitable to be sides of triangle

My idea includes application of stars and bars theorem  to solve
$ a+b+c=2019$  where a, b, c are positive  numbers
and the result will be
$1+a'+1+b'+1+c' = 2019$
$a'+b'+c'=2016$
By stars and bars
Number of triples will be
$2018 C 2$
but i didn't know how can i involve the other conditions to exclude the unappropriate cases
 A: From the second constraint, let $p$ be the constant difference.  Then your terms are $\{a, a+p, a+2p\}$.
We have $$3a+3p=2019\iff a+p=673\iff \boxed {p=673-a}$$
Now, the final constraint is equivalent to $$a+(a+p)>a+2p\iff 2a+p>a+2p\iff a>p\iff a>673-a\iff \boxed {2a>673}$$
Can you finish from here?
A: $\color{blue}{1-)}$ $a ,b,c$ form increasing arithmetic progression , so $a<b<c$
Let say $b=a+x$ , and $c=b+x$  where $x$ is positive integer

*

*$c=a+x+x$


*$b=a+x$
$a+b+c =3a+3x$ where $a,x$ all are positive integers
$\color{blue}{2-)}$ $a , b$ and $c$ are suitable to be sides of triangle , then

*

*$a+b > c \rightarrow 2a+x > a+2x \rightarrow a>x$


*$a+c >b \rightarrow 2a+2x >a+x \rightarrow a >-x$


*$b+c >a \rightarrow 2a+3x > a \rightarrow a>-3x$
Lets say $a=x+k$ where $k$ is positive integer
$a+b+c =6x+3k=2019$ where $k,x$ all are positive integers. Then i lets use generating functions such that

*

*$3k=3,6,9,12,15,...=\frac{x^3}{1-x^3}$


*$6x=6,12,18,24,30,...=\frac{x^6}{1-x^6}$
Now , find $$[x^{2019}]\bigg(\frac{x^3}{1-x^3}\bigg)\bigg(\frac{x^6}{1-x^6}\bigg)$$
