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?