How many 9-digit numbers are there in the decimal system that are divisible by 11 and in which every digit except zero occurs?
Solution. The digits are: $1,2,3,4,5,6,7,8,9$, and we know that 11 | $\overline{a b c d e f g h i}$. Due to divisibility by 11, the digits of $\overline{a b c d e f g h i}$, when summed with alternating signs, result in a number divisible by 11. Therefore, from the given digits, we need to find sums that are divisible by 11 by taking 4 digits as positive and 5 as negative (or 5 as positive and 4 as negative).

In the largest case: $9+8+7+6+5-4-3-2-1=25$.

In the smallest case: $1+2+3+4-5-6-7-8-9=-25$.

Thus, the following numbers divisible by 11 can be considered as sums: $-22,-11,0,11,22$. Since an odd number of odd numbers cannot result in an even number through addition and subtraction, only two sums need to be examined: -11 and 11.

If all digit signs are +, the sum is: $1+2+\ldots+9=45$, from which the sum of the negative-signed digits must be subtracted twice to get -11 or 11.

If the sum of the negative-signed digits is $\frac{45-(-11)}{2}=28$, then for four numbers, the possibilities are:

$$
9+8+7+4, \quad 9+8+6+5
$$

These are the only possibilities because smaller numbers would result in a sum less than 28.

For five numbers, the possibilities are:

$$
\begin{array}{lll}
9+8+7+3+1, & 9+8+6+4+1, & 9+8+6+3+2 \\
9+8+5+4+2, & 9+7+6+5+1, & 9+7+6+4+2 \\
9+7+5+4+3, & 8+7+6+5+2, & 8+7+6+4+3
\end{array}
$$

These are the only possibilities because smaller numbers would result in a sum less than 28.

If the sum of the positive-signed digits is 28, we would get the same 11 arrangements. Therefore, there are 11 ways to assign the digits to positive or negative positions, but they also need to be arranged in order. The five identical signs can be arranged in 5! ways, and the remaining four different signs in 4! ways. Thus, the total number of cases is 11$\cdot$5!$\cdot$4! = 31680.

Therefore, there are a total of 31680 nine-digit numbers in the decimal system that are divisible by 11, where every digit except 0 appears.