Question

Seven digits from the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are written down non-repeatedly in a random order to form a seven digit number. The probability that this seven digit number is divisible by 9 is

• A. $\frac{2}{9}$
• B. $\frac{7}{36}$
• C. $\frac{1}{9}$
• D. $\frac{7}{12}$

Answer 1

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

$n\left(S\right)=$ ${}^9{C}_7$ since the sum of the given $9$ numbers is already $45$.

For divisibility by $9$ ( Sum of the digits to be divisible by $9$ ), we have to

exclude either $1$ and $8$ or $2,7$ or $3,6$ or $4,5$ (So, sum would be $36$)

$n\left(E\right)=4$ (Ways to exclude numbers to get the desired $7$ digits)

$p=\frac{4}{9C_7}=\frac{4\times 7!}{9!}\ast 2!=\frac{1}{9}$