= Solution
If $C=0$, every bin contains at most one ball, so the total number of balls is at most $m$. Since $n=dm>m$, the <pigeonhole principle> makes that event impossible:
$$
\boxed{\mathbb P(C=0)=0.}
$$
As $n\to\infty$ along these values, $m\to\infty$ and the mean occupancy $n/m$ is $d$. Apply part (b):
$$
\frac{\mathbb E[C]}m=1-\left(1-\frac1m\right)^{dm}-d\left(1-\frac1m\right)^{dm-1}.
$$
Both powers tend to $e^{-d}$. Hence
$$
\boxed{\frac{\mathbb E[C]}m\longrightarrow1-(1+d)e^{-d}.}
$$
The <Poisson limit for occupancy fractions> interprets the answer as the probability that a <Poisson distribution> variable with mean $d$ is at least two.
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