= Solution
Under independent assignment,
$$
\mathbb P(Z=z)=\prod_{i=1}^n\pi_{z_i}
=\prod_{r=0}^2\pi_r^{N_r(z)}.
$$
The arm-count vector has a <multinomial distribution>. Conditional on $N_r=n_r$ for $r=0,1,2$, every compatible vector has the same factor $\prod_r\pi_r^{n_r}$, so
$$
\mathbb P(Z=z\mid N_0=n_0,N_1=n_1,N_2=n_2)
=\frac{n_0!n_1!n_2!}{n!}
$$
for compatible $z$, and zero otherwise. Conditioning independent assignment on its arm sizes therefore recovers <complete randomization>.
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