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$