Seven digits from the numbers 1,2,3,4,5,6,7,8 and 9 are written in random order. If the probability that this seven-digit number is divisible by 9 is <img alt="\lambda" src="https://entrancecorner.

Seven digits from the numbers 1,2,3,4,5,6,7,8 and 9 are written in random order. If the probability that this seven-digit number is divisible by 9 is $\lambda$ , then the value of must be
Option: 1
$1999$

Option: 2
$9999$

Option: 3
$1$

Option: 4
$1111$

Answers (1)

Let $\mathrm{a_1, a_2, a_3, a_4, a_5, a_6, a_7}$ be the seven digits and the remaining two be $\mathrm{a_8 \: \: and\: \: a_9.}$
Let $\mathrm{a_1+a_2+a_3+a_4+a_5+a_6+a_7 =9 k, k \in I }$ .........(i)

Also, $\mathrm{a_1+a_2+a_3+a_4+\ldots+a_9 =1+2+3+4+\ldots+9 }$

$\mathrm{ =\frac{9 \times 10}{2} }$

$\mathrm{=45}$ ......................(ii)

Subtracting Eq. (i) from (ii), then

$\mathrm{ a_8+a_9=45-9 k }$ ..............(iii)

Since, $\mathrm{ a_1+a_2+a_3+\ldots+a_9\: and \: a_1+a_2+\ldots+a_7 }$ are divisible by 9 if and only if $\mathrm{ a_8+a_9 }$ is divisible by 9 . Let $\mathrm{ S }$ be the sample space and $\mathrm{ E }$ be the event that the sum of the digits $\mathrm{ a_8 \: \: and\: \: a_9 }$ is divisible by 9 .