Forming Divisible Numbers

A number is divisible by 9 iff its digit sum is divisible by 9. Since 1+2+3...+9=45 is divisible by 9, in order for this number to be divisible by 9 the two numbers that are NOT picked must be also divisible by 9 .

There are 9*8/2=36 different ways to pick two numbers from 9. There are four pairs of digits which add up to 9. 4/36=1/9 . 1+9=10

The total number of numbers are $\left( \begin{matrix} 9 \\ 7 \end{matrix} \right) 7!$

Sum of all the numbers is 45. Now we have to remove 2 numbers such that the sum of remaining 7 numbers is divisible by 9. This is only possible when the sum of the 2 removed numbers is divisible by 9. We cannot achieve a sum of 18 by choosing any two numbers from the given set of numbers as maximum possible sum is 9+8=17 . So, we have to find number of pairs which sum up to 9.

The possible pairs are (8,1),(7,2),(6,3) and (5,4) .So we can remove 2 numbers in 4 ways. And the rest 7 numbers can be arranged in 7! ways.

Hence, probability is $\frac { 4\times 7! }{ \frac { 9!\times 7! }{ 2!7! } }$ which is simply $\frac { 1 }{ 9 }$