= Solution
Express the three counts as sums of <indicator variables> for $B_i=0$, $B_i=1$ and $B_i\geq2$. The <linearity of expectation> applies even though the bin counts are dependent. The <binomial distribution> gives
$$
p_0=(1-1/m)^n,\qquad p_1=\frac nm(1-1/m)^{n-1}.
$$
Thus
$$
\boxed{\mathbb E[E]=m(1-1/m)^n,\qquad \mathbb E[S]=n(1-1/m)^{n-1},}
$$
$$
\boxed{\mathbb E[C]=m\left[1-(1-1/m)^n-\frac nm(1-1/m)^{n-1}\right].}
$$
The relation $E+S+C=m$ checks the sum of these <expectations>. For $m=1$, use the direct deterministic occupancies, or the usual binomial convention $0^0=1$ when $n=1$.
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