My math question

Discussion in 'Scientific Statistics Math' started by PortlandDream, Dec 30, 2011.

  1. Hello,

    I was wondering if there might be a formula to solve this problem I
    devised.

    A = number of people
    B = number of apples
    C = total possible variations

    Where you can assign to all values of A all different values of B (so
    that the sum always ends up being B) to get the total variations.

    I was kinda dreaming about an apple a day and thought if I could
    spread it out, or if someone hogged it all plus all variations. Yeah I
    am a math nerd.

    I thought I have seen a formula for this, but I cannot recall it for
    the life of me.

    I have made some tables for this but it is clear the system will
    spiral out of hand without a formula and I cannot derive one so far
    with the information at hand.

    Any help would be awesome!
     
    PortlandDream, Dec 30, 2011
    #1
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  2. PortlandDream

    Rich Ulrich Guest

    Obsessing about combinations has never been one of my
    hobbies, but here are a few comments.

    Sometimes the solutions to combinatorial problems seem to
    "spiral out of hand" even when you have a formula -- if that
    means "a lot of tedious detail" when you are not talking about one
    of the extreme cases.

    For a better chance for help with this particular problem --
    You need to be more specific about what you want to count, the
    detailed layout of the problem. I *think* that you want to
    distribute B indistinguishable apples among A indistinguishable
    people; but otherwise, it is a different problem. Are you talking
    about small numbers, or *very* small numbers, or something else?
    - If person #1 has all the apples, is that same solution as if
    person #2, or #3, or ..., has all the apples?

    Tedious detail:
    For 5 apples and over 5 people, your answer set includes
    (5,...), (4,1,...), (3,1,1,...), (3,2, ...), (2, 1,1,1, ...), (2,2,1),
    (1,1,1,1,1).

    If the people have labels, there more answer sets, and
    you are probably interested in a tiny subset of possible
    questions.

    Hope this helps.
     
    Rich Ulrich, Jan 1, 2012
    #2
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  3. Umm, the hardest version with each person named.

    So if three apples, three people, for example, we would see 9 results

    (3 . .) (. 3 .) (. . 3) (2 1 .) (2 . 1) (1 2 .) (. 2 1) (1 . 2) (1 1
    1)

    If I just could remember the name of the method I could sleep in
    peace :(
     
    PortlandDream, Jan 2, 2012
    #3
  4. PortlandDream

    Rich Ulrich Guest

    Or, it may be considered simpler to list those answers
    because you don't have to find the "similar" cases
    that get added together.
    Combinatorials.
    Combinatorics formulas.
     
    Rich Ulrich, Jan 2, 2012
    #4
  5. Borrow two oranges and prepare a row of three plus two positions. In
    each position, put either an apple or an orange. Every way of doing
    this creates three groups of apples, separated by the oranges.

    The number of the arrangements is the number of ways to choose the
    positions for the oranges.
    No name comes to mind.
     
    Jussi Piitulainen, Jan 2, 2012
    #5
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