I need to find the solution of $\dfrac{4^n}{n!}=51.8682$. Though I need $n$ to be a natural number, if $n\in\mathbb{R}$, I can take the celling of $n$ for problem purpose.
It can be done by trial and error method. But one thing is noticeable that the denominator is increasing rapidly than the numerator. For $n=1,2,3$ the R.H.S is very far from the result. As $n$ increases, the fraction decreases and hence the fraction does not reach even at $50$.
Is there any way to solve this equation?
Question added A cafeteria with self-service has an arrival rate of $12$/hour. The average time taken by a person to collect and eat his meal is $20$ minutes. Assuming that the inter-arrival times are exponentially distributed, how many seats must the cafeteria have to accommodate each customer with probability $0.95$?
Solution: Here $\lambda=12$ persons/hour and $\mu=\dfrac{60}{20}=3$ customers/hour. Therefore $\rho=\dfrac {12}{3}=4$. We need to find $n$ so that $P_n=0.95$. Therefore we have $ \dfrac{\rho^ne^{-\rho}}{n!}=0.95\implies \dfrac{4^ne^{-4}}{n!}=0.95\implies \dfrac{4^n}{n!}=51.8682$
Working in the real domain, you can have an almost exact explicit solution.
Making the problem more general, you want to solve for $n$ $$\frac {a^n}{n!}=b \qquad \implies \qquad n!=\frac {a^n}{b}$$
If you look at this old question of mine, you will see a superb approximation by @robjohn (one of our moderators).
Just make $10^k=\frac 1b$ that is to say $k=-\frac {\log(b)}{\log(10)}$ to get $$n \sim e a \exp\left(W\left( -\frac{\log \left(2 \pi a b^2\right)}{2 e a} \right)\right)-\frac 12$$ where $W(.)$ is Lambert function
The problem is that, with your numbers $(a=4,b=51.8682)$, the solution is a complex number.
Changing $b$ to its reciprocal would give $n=12.5120$ while its exact value is $12.5147$.