An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5, 6,7,8,9 without replacement. The number of ways in which this can be done is

Question

An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5, 6,7,8,9 without replacement. The number of ways in which this can be done is (A) 9 ! (B) 2(7!) (C) 4(7 !) (D) (36) (7!)

Tardigrade Admin · Accepted Answer

We have 0+1+2+3 ldots+8+9=45 To obtain an eight digit number exactly divisible by 9 , we must not use either (0,9) or (1,8) or (2,7) or (3,6) or (4,5). [Sum of the remaining eight digits is 36 which is exactly divisible by 9 .] When, we do not use (0,9), then the number of required 8 digit numbers is 8 !. When one of (1,8) or (2,7) or (3,6) or (4,5) is not used, the remaining digits can be arranged in 8 !- 7 ! ways. 0 cannot be at extreme left. Hence, there are 8 !+4(8 !-7 !)=(36)(7 !) numbers in the desired category.

Q. An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits $0, 1, 2, 3, 4, 5$ , $6, 7, 8, 9$ without replacement. The number of ways in which this can be done is

$9!$

$2 (7!)$

$4 (7!)$

$(36) (7!)$

Q. An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5, 6,7,8,9 without replacement. The number of ways in which this can be done is

9!

2(7!)

4(7!)

(36)(7!)

Q. An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits $0, 1, 2, 3, 4, 5$ , $6, 7, 8, 9$ without replacement. The number of ways in which this can be done is

$9!$

$2 (7!)$

$4 (7!)$

$(36) (7!)$