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Math Help - Combinatorial estimate using Stirling's formula

  1. #1
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    Question Combinatorial estimate using Stirling's formula

    Hey,

    Could anyone please help with the following question?

    "Estimate the number of decimal digits in nC(n/3) as a function of n, where n is divisible by 3"

    Using Stirling's formula as n! ~ (sqrt[](2*pi*n))*((n/e)^n)*O(n) where O(n) is the error.

    I substituted and cancelled the equation down to this:
    nC(n/3) = n! / (2n/3)!(n/3)! = .... = (3^n)O(n) / (2^n)(sqrt[]pi*n)O(2n/3)O(n/3)

    And this is irrational therefore I know I'm wrong but can't find my error.

    Thanks a bunch!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DaRush19 View Post
    Hey,

    Could anyone please help with the following question?

    "Estimate the number of decimal digits in nC(n/3) as a function of n, where n is divisible by 3"

    Using Stirling's formula as n! ~ (sqrt[](2*pi*n))*((n/e)^n)*O(n) where O(n) is the error.

    I substituted and cancelled the equation down to this:
    nC(n/3) = n! / (2n/3)!(n/3)! = .... = (3^n)O(n) / (2^n)(sqrt[]pi*n)O(2n/3)O(n/3)

    And this is irrational therefore I know I'm wrong but can't find my error.

    Thanks a bunch!
    That is not very clear, but being irrational is not important (its an approximation), what you want is the \log_{10} of this plus one for the approximate number of decimal digits.

    CB
    Last edited by CaptainBlack; August 30th 2009 at 05:27 AM.
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  3. #3
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    log_{10}(3^n / 2^n\sqrt{n\pi})?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by DaRush19 View Post
    log_{10}(3^n / 2^n\sqrt{n\pi})?
    Well my notes are unintelligable, but it looks wrong as the 2 and 3 should have different exponents. So I have run this through Maxima to get:

    Combinatorial estimate using Stirling's formula-ncnb3.png

    Also you have to add 1 from the log to get the number of digits (and round down to the nearest integer)

    CB
    Last edited by CaptainBlack; August 30th 2009 at 05:28 AM.
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