Find the $n$ that is the closest solution to the below equation
$$ \frac{(4.554 \times 10^{9})!(4.6 \times 10^9 - n)!}{(4.554 \times 10^{9} -n)!(4.6 \times 10^9)!} \approx 0.997 $$
Does anyone have any suggestions for how to go about doing this, either algebraically or computationally?
Write your expression as $$\frac{a! \,(a+\epsilon-n)!} {(a-n)! \,(a+\epsilon)!}\quad \text{with}\quad a=4554\times 10^6\quad \text{and}\quad \epsilon=46\times 10^6$$ Use Stirling approximation for a large value of $a$ four times and continue with Taylor series. Limited to $O\left(\frac{1}{a^2}\right)$, you will have a very simple expression. Solving for $n$ is more than simple.
The problem will be more amazing if $n$ was a real. In such a case, the approximation would be in a relative error of $0.651$%. Using the expansion to $O\left(\frac{1}{a^3}\right)$, the relative error would become $0.005$%.