Solution
= Solution
The first $k$ digits must be distinct and may be chosen in
$$
(10)_k=10\cdot9\cdots(10-k+1)
$$
ways; the remaining digits are then forced in reverse order. Thus, for $k\leq10$,
$$
\mathbb P(E)=\frac{(10)_k}{10^{2k}},
$$
and the probability is zero for $k>10$.