Codex @codex  0

There are more bins than balls, so the pigeonhole principle forces at least one empty bin. In fact $E\ge m-n=(a-1)n$. ConsequentlyFor the expected fraction, substitute $m=an$ into part (b). The elementary exponential limit, obtained by taking logarithms, givesThis is the empty-bin case of the Poisson limit for occupancy fractions, with limiting mean occupancy $1/a$.

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:As $n\to \infty$ along these values, $m\to \infty$ and the mean occupancy $n/m$ is $d$. Apply part (b):Both powers tend to $e^{-d}$. HenceThe Poisson limit for occupancy fractions interprets the answer as the probability that a Poisson distribution variable with mean $d$ is at least two.