1. Call a positive integer a "random number" if and only if
(1) all digits are non-zero;
(2) the number is divisible by 11;
(3) the number is divisible by 12, and any permutation of its digits also results in a number divisible by 12.
Ask: How many ten-digit random numbers are there?
1. From condition (3), we know that the digits of the random number can be arbitrarily permuted, and the number remains divisible by $12 = 3 \times 4$ after rearrangement. From condition $(1)$, we know that the digits of this random number can only be $2, 4, 6,$ and $8$.
Next, we show that the random number can only contain the digits 4 and 8.
In fact, assume the random number contains the digit 2. If the random number is rearranged such that the last digit is 2, the last two digits could be $22, 42, 62,$ or $82$. In this case, the new random number cannot be divisible by 4 (and thus not by 12), which contradicts property (3).
Similarly, the random number cannot contain the digit 6.
Let the ten-digit random number have $k$ digits of 8 and $10-k$ digits of 4, where $0 \leqslant k \leqslant 10$.
Then the sum of the digits of the ten-digit random number is
$$
8k + 4(10 - k) = 40 + 4k = 36 + 4(k + 1).
$$

Note that a random number is divisible by 3 if and only if the sum of its digits is divisible by 3.
Thus, $3 \mid (k + 1) \Rightarrow k = 2, 5, 8$.
A random number is divisible by 11 if and only if the sum of the digits in the even positions equals the sum of the digits in the odd positions.
We consider the cases for $k$.
(i) $k = 2$.
In this case, one 8 is in an odd position, and the other 8 is in an even position, giving $5 \times 5 = 25$ methods.
(ii) $k = 5$.
Since 8 cannot be divided into two equal groups, there is no solution in this case.
(iii) $k = 8$.
There are 4 digits of 8 in the odd positions and 4 digits of 8 in the even positions, giving $5 \times 5 = 25$ methods.

Therefore, the number of ten-digit random numbers that satisfy the conditions is $25 + 25 = 50$.