South-Afrika, 2011

If $a_i = b_j = 0$, then $c_i = c_j = 0$. If $(c_1, c_2, \dots, c_p)$ is a rearrangement of $(0, 1, \dots, p-1)$, then no entry can appear more than once. Hence $i = j$.

Without loss of generality we may suppose that $a_p = b_p = 0$ so that $c_p = 0$. Then we have
$$(p-1)! \equiv_p a_1 a_2 \dots a_{p-1} \equiv_p b_1 b_2 \dots b_{p-1} \equiv_p c_1 c_2 \dots c_{p-1}$$
and so
$$\begin{align*} (p-1)! &\equiv_p c_1 c_2 \dots c_{p-1} \\ &\equiv_p (a_1 a_2 \dots a_{p-1}) (b_1 b_2 \dots b_{p-1}) \\ &\equiv_p ((p-1)!)^2, \end{align*}$$
which implies that $(p-1)! \equiv_p 1$. However, since $p$ is odd, this contradicts Wilson's Theorem, and we conclude that $(c_1, c_2, \dots, c_p)$ is not a rearrangement of $(0, 1, \dots, p-1)$.