• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Combinatorics "Piles" (1 Viewer)

frog1944

Member
Joined
Apr 9, 2016
Messages
210
Gender
Undisclosed
HSC
2017
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
 

shehan123

Member
Joined
Aug 3, 2016
Messages
30
Gender
Undisclosed
HSC
N/A
The reason you ever divide by the number of groups factorial is because choosing

ABC DEF GHI

is the same as

DEF GHI ABC

therefore we get 3! more arrangements.

in the 2nd situation, there is an unequal number of people in each of the groups, therefore you dont need to divide by 3! as such.

Meaning, there is no way those (n+1) people can be put together apart from when they're in the group of n+1 people, they cannot be put together in a group of (n-1) people because 2 wont fit.
Think of it as arranging marbles, if i have 5 distinct marbles : 5!
but if i have 5 marbles, 2 of which are identitical: 5!/2!

Likewise, if theres two group sizes that are identical, or all group sizes are identical, you divide it by (the number of identical groups, factorial). Hope it makes sense, i probably havent answered it properly.
 
Last edited:

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
I thought this thread was going to be about a medical condition caused by spending too much time contemplating Combinatorics in the bathroom.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top