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.

since $10 = -1 \pmod {11}$, a number $abcdefghi$ is a multiple of $11$ if and only if $(a+c+e+g+i)-(b+d+f+h)$ is a multiple of $11$.
Since $(a+c+e+g+i)+(b+d+f+h) = 45 = 1 \pmod {11}$, this is equivalent to $1-2(b+d+f+h) = 0 \pmod {11}$, and to $(b+d+f+h) = 6 \pmod {11}$.
So we want to know, when is the sum of $4$ numbers randomly taken in $\{1 ; \ldots ; 9 \}$ is congruent to $6$ modulo $11$.
Clearly, if we were picking our four numbers in $\{1 ; \ldots ; 11\}$ (or $\{0 ; \ldots ; 10 \}$), the sum is uniformly distributed mod $11$ (if we add $3$ to each number, it's like adding $1$ to the sum). Which means there are $\frac 1 {11}\binom {11}4 = 30$ good quadruplets there.
Out of all of those we are only interested in those that don't use $10$ nor $11$.
Let's count how many use $10$ : 
A quadruplet that use $10$ and that sums to $6$ is $10$ plus a triplet that sums to $7$ and that doesn't use $10$.
Once again we count the total number of triplets that sum to $7$, but we again have extra triplets, those that contain $10$.
We can continue like this, to remove the extra triplets we have to remove pairs that sum to $8$, and finally remove from those pairs the pair $\{10 ; 9 \}$.
So we get $\frac 1 {11}(\binom {11}3 - \binom {11}2 + \binom {11}1) = 11$ quadruplets that sum to $6$ and use $10$, which means there are $\frac 1 {11}(\binom {11}4 - \binom {11}3 + \binom {11}2 - \binom {11}1) = 19$ quadruplets that sum to $6$ and don't use $10$.
Now, we count the quadruplets that use $11$. The same thing happens the same way, even at the last step (because $4 \times 11 \neq 6 \neq 4 \times 10)$.
Had we wanted to count the number of quadruplets that don't use $7$, then we would have a difference at the end (because $7 \times 4 = 28 = 6$): none of the pairs that sum to $3$ contained a $7$ in the first place, so we don't count that last $\frac 1{11}\binom {11}1$.
Or said another way, the sum of quadruplets that don't use $10$ is almost uniform : it hits every sum $19$ times except $4 \times 10 = 7$, who is hit $20$ times (for a total of $210$, and there are $210$ quadruplets that don't use $10$).
Finally we want to count how many quadruplets sum to $6$ and use both $10$ and $11$. Those are the number of pairs that sum to $7$ and don't use $10$ nor $11$.
There are $5$ pairs that sum to $7$, one of which uses $10$ and one of which uses $11$ (none use both because $10+11 \neq 7$)
So that's a total of $3$ quadruplets that sum to $6$ and use both $10$ and $11$.
The final number is $30 - 11 - 11 + 3 = 11$ quadruplets that sum to $6$ and don't use $10$ or $11$.
Since there are $126$ quadruplets that don't use $10$ or $11$, the final probability is $\frac {11}{126}$
