If I have $r$ balls and $n$ urns and each ball is randomly put in an urn with equal probability then the probability of the first urn having $k$ balls is
$${r \choose k}\frac{(n-1)^{r-k}}{n^r}.$$
I have to show that as $r$ and $n$ tend to infinity but $\dfrac{r}{n}$ is constant the probability tend to ${e^{-m}}\dfrac{m^k}{k!}$. I don't know how to do this part. This question came from a book on probability and it doesn't teach analysis.
\begin{align} & {r \choose k}\frac{(n-1)^{r-k}}{n^r} \\ &= \frac{1}{k!} \; \frac{r(r-1) \cdots (r+1-k)}{n^k} \; \frac{(n-1)^{r-k}}{n^{r-k}} \\ &= \frac{1}{k!} \; \frac{r}{n} \; \frac{r-1}{n} \cdots \frac{r+1-k}{n} \left(1-\frac1n\right)^{r-k} \end{align}
For any $k$ fixed, let $m = \dfrac{r}{n}$ be a constant. As $r,n \to \infty$ with $m$ fixed, the fractions in the middle $\dfrac{r+1-j}{n}$ tends to $m$ for each $j \in \{1,\dots,k\}$.
The factor on the right: $$\left(1-\frac1n\right)^{r-k} = \left(\left(1-\frac1n\right)^{n}\right)^m \; \left(1-\frac1n\right)^{-k} \to e^{-m} \cdot 1$$
Therefore, the product tends to ${e^{-m}}\dfrac{m^k}{k!}$.