Question

An eight digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is:

• A. $72\left(7!\right)$
• B. $18\left(7!\right)$
• C. $40\left(7!\right)$
• D. $36\left(7!\right)$

Answer 1

Answer: (D)

$36\left(7!\right)$

If no is divisible by $9$ then out of all number

those number whose digit sum is divisible by

$9$ will be divisible by $9$

Now is $8$ digits are to be chosen from digits that means $2$ digits will not be selected

If those $2$ digits are

$\begin{array}{c}0,9Sumofdigits=36\\ 1,8Sumofdigits=36\\ \begin{array}{c}2,7Sumofdigits=36\\ \begin{array}{c}3,6Sumofdigits=36\\ 4,5Sumofdigits=36\end{array}\end{array}\end{array}\}Divisibleby9$

no other case will gice us a sum of digit divisible by $9$

Total ways to do this $=8!+7!7+7!7+7!7+7!7$

$=36\times 7!$