Solution:
A number is divisible by 11 only if the difference of the sum of the digits at odd places x and sum of the digits at even places y is divisible by 11 i.e., 0, 11, 22, 33 ...
Here, the sum of all the 9 digits (1, 2, 3, ...9) is 45.
We cannot create the difference of zero
since x + y = 45, which is odd hence cannot be broken into two equal parts integers.
Now, we will took for the possibilities of 11 for which set of even numbers are as follows:
{1, 2, 6, 8} (1, 2, 5, 9) {1, 3, 6, 7}
{1, 3, 5, 8} {1, 3, 4, 9} {1, 4, 5, 7}
{2, 3, 5, 7} {2, 3, 4, 8} {2, 4, 5, 6}
and {4, 7, 8, 9} {5, 6, 8, 9}
The above set of values either gives the sum of 17 or 28. Since if the sum of 4 digits at even places be 17 or 28 then the sum of rest of the digits (i.e., digits at odd places) be 28 or 17 respectively and thus we can get the differences of 11. Further we cannot get the difference of 22 or 33... So there is only possible difference that can be created is 11 and there are only 11 set of values given above containing 4! ways and the remaining 5 digits can be arranged in 5! ways.
Thus the favourable number of numbers = 11 × 4! × 5!
But the total number of ways of arranging a nine digit number is
= 9!
∴ Exclusive number of cases = 9!
∴ Required probability =
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