Question

The seven digit numbers divisible by $9$ formed with digits $\{1,2,3,4,5,6,7,8,9\}$. The number of ways in which this can be done is

• A. $7!$
• B. $2.\left(7\right)!$
• C. $3.\left(7\right)!$
• D. $4.\left(7\right)!$

Answer 1

Answer: (D)

$4.\left(7\right)!$

The sum of the all given digits is 45. The number is divisible by $9$ if sum of digits is divisible by $9.$
So the seven digit number should not contain $(1,8)$ or $(2,7)$ or $(3,6)$ or $(4,5)$ and the remaining digits can be arranged in $7!$ ways
$\therefore$ Number of such numbers is $=4(7!)$