Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$
I know the condition for divisibility by $11$ but I couldn't guess how to apply it here.
Please help me in this regard. Thanks.

A number is divisible by 11 if and only if the sum of the odd-position digits and the even-position digits differ by a multiple of 11. Now, the sum of the digits from 1 to 9 is odd, so the difference in any such number must be either 11 or 33. The only subsets of $\{1,2,3,4,5,6,7,8,9\}$ where the difference between the subset and its complement is $33$ have six or more elements; this is impossible, since there are 5 odd-position digits and 4 even-position digits. Thus the two sets of digits differ by 11, and then the larger sum is $28$ and the smaller is $17$.
So we want collections of 4 or 5 integers from $\{1,2,3,4,5,6,7,8,9\}$ whose sum is $28$. There are eleven such (it's pretty straightforward to simply enumerate these by hand):
\begin{align*}
   &\{4, 7, 8, 9\}, \{5, 6, 8, 9\}, \{1, 3, 7, 8, 9\}, \{1, 4, 6, 8, 9\}, \\
   &\{1, 5, 6, 7, 9\}, \{2, 3, 6, 8, 9\}, \{2, 4, 5, 8, 9\}, \{2, 4, 6, 7, 9\}, \\
   &\{2, 5, 6, 7, 8\}, \{3, 4, 5, 7, 9\}, \{3, 4, 6, 7, 8\}.
\end{align*}
Each of these yields $24\cdot 120$ possible numbers (arrange the 5-set any way you want, and the 4-set any way you want), for a total of $11\cdot 24\cdot 120 = 31680$ possibilities. (Note that this corresponds roughly to the numeric estimate of $0.08$ given in the comments).
