Hi,
I'm trying to figure out what I'm doing wrong with a question.
Context: If I wanted to arrange 12 presents into 4 piles of 3, this can be done in (12C3*9C3*6C3*3C3)/(4!) as the piles are interchangeable (for the last part about 4!).
Question: However, if I wanted to arrange 3n students into 3 groups of n-1, n+1 and n students, this is (3nCn*(2n+1)Cn), though why is there no 3! on the denominator? I understand that if the groups aren't interchangeable, but they are aren't they? If not, why not?
Thanks
I'm trying to figure out what I'm doing wrong with a question.
Context: If I wanted to arrange 12 presents into 4 piles of 3, this can be done in (12C3*9C3*6C3*3C3)/(4!) as the piles are interchangeable (for the last part about 4!).
Question: However, if I wanted to arrange 3n students into 3 groups of n-1, n+1 and n students, this is (3nCn*(2n+1)Cn), though why is there no 3! on the denominator? I understand that if the groups aren't interchangeable, but they are aren't they? If not, why not?
Thanks