P&C-2

For a number to be divisible by 9,the sum of the digits should be divisible by 9.

Case 1:(if there is no 0 in the number )

Sum of digits = 45 (divisible by 9)

No. of possible arrangements of the digits = 9!.

Case 2:(if the no. has a 0)

0 must replace some digit.If it replaces any digit other than 9 then the sum does not remain divisible by 9.

Hence the number does not contain 9 in this case.

0 can be placed in any of the 8 positions (because 0 cannot be placed as the first digit)

The rest of the 8 digits can be placed in 8! ways.

\thereforeTotal no. of digits = 9!+8*8!=17*8!

k=17

17