# Math Help - Combinatorial estimate using Stirling's formula

1. ## Combinatorial estimate using Stirling's formula

Hey,

"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!

2. Originally Posted by DaRush19
Hey,

"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

3. $log_{10}(3^n / 2^n\sqrt{n\pi})$?

4. Originally Posted by DaRush19
$log_{10}(3^n / 2^n\sqrt{n\pi})$?
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:

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

CB