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Thread: Combinatorics "Piles"

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    Combinatorics "Piles"

    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
    Mathematics Extension 2 - Physics - Chemistry - Economics - English Advanced

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    Re: Combinatorics "Piles"

    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 by shehan123; 21 Oct 2017 at 1:23 PM.

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    Re: Combinatorics "Piles"

    Oh, that makes sense. Thank you very much!
    Mathematics Extension 2 - Physics - Chemistry - Economics - English Advanced

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    Re: Combinatorics "Piles"

    I thought this thread was going to be about a medical condition caused by spending too much time contemplating Combinatorics in the bathroom.
    pikachu975 likes this.

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