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Number of possible unique outcomes help - Binomial

  1. Mar 24, 2009 #1
    Hi,

    My question is based around the idea about calculating the number of possible outcomes when a given number of variables are chosen randomly (all with equal probability of being picked) a given number of times. Most importantly, I an specifically working so that order is redundant. ie AAB = ABA = BAA and would therefore only count those 3 possibilities as one possible "unique" outcome.

    I'm aware that if for example AB is not equal to BA in the context of this question then the number of possible outcomes would just be:

    (number of variables)^(number of states or "picks")


    Example: for 3 variables ABC and 3 Picks

    AAA
    AAB
    ...
    ...
    ABC
    ...
    ...
    CCB
    CCC

    out of all 27 possible combinations, there are only 10 unique ones

    From observation and a little work I managed to get the formula:

    f(v,p) = [v(v+1)(v+2)...(v+p-1)]/(p!)

    where: v= number of variables
    p= number of picks

    and it works, but interestingly (to me at least) was that the answer could be taken directly from pascals triangle.

    by p= p'th row on pascals traingle (taking top row = 0)
    v= v'th value on the diagnal from either side

    ...

    I know a bit about binomial theorem, possible outcomes, "nCr" etc but could someone quickly give me a direction/link from the derived formula and pascal/binomial theorem..

    Many Thanks
    Tom
     
  2. jcsd
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