Five different digits from the set of numbers {1, 2, 3, 4, 5, 6, 7} are written in random order. Find the probability that five-digit number thus formed is divisible by 9.
If the sum of the digits in a number is divisible by 9, then the number itself is divisible by 9.
The least and the greatest sum of five digits chosen from the set are 15 and 25, respectively.
So, either number comprises {1, 2, 4, 5, 6} or {1, 2, 3, 5, 7}
`therefore` Number of favourable cases = `2 xx 5!`
Total number of cases = `.^(7)P_(5)`
`therefore` Required probability = `(2 xx 5!)/(.^(7)P_(5))=(2)/(21)`