<s>[INST] Number of solutions, $a+b+c=n$, $a\gt b\gt c\ge0$ Number of non negative integral solutions for $a + b + c = n$
Where $n$ is a positive integer are
$$\binom{n + 3 - 1}{3 - 1}$$
But if a condition is there $a > b > c$
Is there any direct method by which we can find out required number of solutions.
I believe that we should multiply original number of solutions by $1 / 4$ as there are four following cases possible
$a = b > c$
$a = b = c$
$a > b = c$
$a > b > c$
 [/INST] When $c=0$, there are $[(n-1)/2]$ solutions, since $b$ can be any positive integer not exceeding $(n-1)/2$. 
When $c\ne0$, let $d=a-c$, $e=b-c$, then $d\gt e\gt0$ and $d+e=n-3c$, so the number of solutions (by the previous argument) is $[(n-3c-1)/2]$. 
$c$ can be anything from 0 to $[n/3]-1$, so the answer is $$f(n)=\sum_{c=0}^m\left[{n-3c-1\over2}\right]$$ where $m=[n/3]-1$. Now we need a closed form for $f(n)$. Here's one way to go. Note $$f(n+6)=\sum_{c=-2}^{m}\left[{n-3c-1\over2}\right],\qquad f(n+12)=\sum_{c=-4}^{m}\left[{n-3c-1\over2}\right]$$ so $$f(n+12)-2f(n+6)+f(n)=\left[{n+11\over2}\right]+\left[{n+8\over2}\right]-\left[{n+5\over2}\right]-\left[{n+2\over2}\right]=6$$ Thus on any arithmetic progression with common difference 6, $f(n) $ is a quadratic polynomial. So we just have to find the 6 quadratics. 
We calculate $f(3)=1$, $f(9)=7$, $f(15)=19$, so (by standard techniques) $f(6n+3)=3n^2+3n+1$; the whole deal looks like this: $$\eqalign{f(6n)&=3n^2\cr f(6n+1)&=3n^2+n\cr f(6n+2)&=3n^2+2n\cr f(6n+3)&=3n^2+3n+1\cr f(6n+4)&=3n^2+4n+1\cr f(6n+5)&=3n^2+5n+2\cr}$$ A simpler formula is $$f(n)=\left[{n^2+6\over12}\right]$$
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