$\star$ Suppose that $n$ balls are tossed into $n$ bins, where each toss is independent and the ball is equally likely to end up in any bin. What is the expected number of empty bins? What is the expected number of bins with exactly one ball?

Both of them are $n/e$ or at least are approximatelly close to it when $n$ is large enough. Let's do empty bins first.

Let $X_i$ be the event that all the balls fall in bins, other than the $i$th.

$$ \Pr\{X_i\} = \bigg(\frac{n-1}{n}\bigg)^n = \bigg(1 - \frac{1}{n}\bigg)^n \approx \frac{1}{e} $$

The expectation:

$$ \E[X] = \sum_{i=1}^n\E[X_i] = \frac{n}{e} $$

It's quite similar with exactly one ball. The probability is:

$$ \Pr\{Y_i\} = n\frac{1}{n}\bigg(\frac{n-1}{n}\bigg)^{n-1} = \bigg(\frac{n-1}{n}\bigg)^{n-1} \approx \frac{1}{e} $$

The expectation is the same.

Here's a Math/StackExchange question that clarifies it.