# combinatorial – \$ a + b + c = 2001 \$ and \$ a> b> c> 0 \$ How many ordered triples are possible?

$$a + b + c = 2001$$ Y $$a> b> c> 0$$ How many ordered triples are possible?

I tried to replace these equations.

$$c + x = b$$

$$b + y = a$$

well, the equation would be

$$3c + 2x + y = 2001$$

but I do not know how to solve this, is there a more general formula of stars and bars?