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

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
$$y(y-1)=730\log(2\quad \implies \quad y=\frac{1}{2} \left(1+\sqrt{1+2920 \log (2)}\right)=22.999943$$
