Question

Seven digits from the numbers 1,2,3,4,5,6,7,8 and 9 are written in random order. The probability that this seven digit number is divisible by 9 is:

• A. $\frac{7}{9}$
• B. $\frac{1}{9}$
• C. $\frac{2}{9!}$
• D. $\frac{4}{9}$

Answer 1

The correct option is B $\frac{1}{9}$

Total 7 digit numbers can be formed from the 9 digits = ${}^9{P}_7$
There are four exclusive cases of selecting 7 digits out of 9 digits which can from 7 digit numbers which are divisible by 9.
2,3,4,5,6,7,9} 36 removing 1 and 8
1,3,4,5,6,8,9} 36 removing 2 and 7
1,2,4,5,7,8,9} 36 removing 3 and 6
1,2,3,6,7,8,9} 36 removing 4 and 5
All the 7 numbers of each of the 4 sets can be arranged in 7! ways.
Hence the favourable number of numbers = $4\times 7!$
$\therefore$ Required probability = $\frac{4\times 7!}{{}^9{P}_7}=\frac{1}{9}$.