Solve Algebraical.ly $0.5=\dfrac{365!}{365^{n}(365-n)!} $

Question

How does one go about solving this equation? Not sure how to approach this as no factorials will cancel out. Im sorry I meant $\dfrac{365!}{365^{n}(365-n)!}=0.5$.

Answer 1

It has no solution: $\dfrac{365!}{n!(365-n)!}=\dbinom{365}n$, the number of subsets of $\{1,2,\ldots,365\}$ of cardinality $n$, which is always an integer.

Added: This answer applies to the original version of the question.

Answer 2

We usually solve this equation numerically:

$$a_n=\frac{365!}{365^n(365-n)!}$$

Hence $a_1=1$ and $$a_{n+1}=a_n.\frac{365-n}{365}$$

If you want to solve $a_n=p$, just do a little program that computes $a_n$ from $a_1$ by multiplying at each step by $\frac{365-n}{365}$ until you find $p$.

Here $$a_{23}=0.4927027656$$

Answer 3

/* birthProb_0.cc  25-nov-2013
http://math.stackexchange.com/users/85343/felix-marin

http://math.stackexchange.com/questions/580287/solve-algebraical-ly-0-5-dfrac365365n365-n/580318#580318
*/
#include <cmath>
#include <iostream>
using namespace std;
typedef long double ldouble;
typedef unsigned long long ullong;
const ldouble LN365=log(365.0L);

inline ldouble lnFac(ullong n)
{
 return lgammal(static_cast<ldouble>(n) + 1.0L);
}
const ldouble LN365FAC=lnFac(365ULL);

inline ldouble f(ullong n)
{
 return exp(LN365FAC - n*LN365 - lnFac(365ULL - n));
}

int main()
{
 ullong n=0;
 while (   ( n<=365 ) && ( f(n)>0.5L )  ) ++n;

 (--n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;
 (++n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;
 (++n),cout<<"n = "<<n<<" ---> "<<f(n)<<endl;

 return 0;
}

Result: $$ \begin{array}{l} {\tt n = 22 ---> 0.524305} \\ \color{#0000ff}{\large{\tt n = 23 ---> 0.492703}} \\ {\tt n = 24 ---> 0.461656} \end{array} $$

Answer 4

Suppose that you want to solve in the real domain the equation $$\frac {x!}{x^n (x-y)!}=\frac 12$$ take logarithm to make $$\log(x!)-y \log(x)-\log[(x-y)!]=-\log(2)$$ Using twice Stirling approximation for large values of $x$, we have $$-\log(2)=-\frac{(y-1) y}{2 x}+O\left(\frac{1}{x^2}\right)$$ Ignoring the higher order terms and making $x=365$ gives $$y(y-1)=730\log(2\quad \implies \quad y=\frac{1}{2} \left(1+\sqrt{1+2920 \log (2)}\right)=22.999943$$