What are the number of ways of arranging 30 identical objects into three groups such that no two groups contain the same number of objects.

Question

anonymous · Answer

I'll give you an indication: it's the nomber of solution of the equation a+b+c=30 with a and b and c are differents and, a and b and c are positive integer <=30

zarkon · Answer

If you can have have zero items in a group

$${32\choose 2}-1-3\times(15)=450$$

if you have to have a least one object in each group then it is 


$${29\choose 2}-1-3\times(13)=366$$