We all know that the number of solutions to a+b = 3 (where a and b are non negative) can be easily found out by stars and bars theorem. So answer of the above is 4. { (3,0),(0,3),(1,2),(2,1)}. But I want the answer to be 2 as the first two and last two are equivalent. Formally, as the stars and bars give the number of ordered sets, I want to find out the number of unordered sets.