A seven-digit number without repetition and divisible by 9 is to be formed by using seven digits out of 1,2,3,4,5,6,7,8,9. The number of ways in which this can be done is

Question

A seven-digit number without repetition and divisible by 9 is to be formed by using seven digits out of 1,2,3,4,5,6,7,8,9. The number of ways in which this can be done is (A) 9 ! (B) 2(7 !) (C) 4(7 !) (D) None of these

Tardigrade Admin · Accepted Answer

Sum of 7 digits is a multiple of 9 . Sum of numbers 1,2, 3,4,5,6,7,8,9 is 45 ; so two left digits should also have sum of 9 . The pairs of left numbers are (1,8),(2,, 7), (3,6),(4,5). With each pair left number of 7 -digit number is 7 ! So with all 4 pairs, total number is 4 × 7 !.

Q. A seven-digit number without repetition and divisible by 9 is to be formed by using seven digits out of $1, 2, 3, 4, 5, 6, 7, 8, 9$ . The number of ways in which this can be done is

$9!$

$2 (7!)$

$4 (7!)$

None of these

Q. A seven-digit number without repetition and divisible by 9 is to be formed by using seven digits out of 1,2,3,4,5,6,7,8,9. The number of ways in which this can be done is

9!

2(7!)

4(7!)

None of these

Sum of 7 digits is a multiple of 9 . Sum of numbers 1,2,3,4,5,6,7,8,9 is 45 ; so two left digits should also have sum of 9 . The pairs of left numbers are (1,8),(2,, 7), (3,6),(4,5). With each pair left number of 7 -digit number is 7! So with all 4 pairs, total number is 4×7!.

Q. A seven-digit number without repetition and divisible by 9 is to be formed by using seven digits out of $1, 2, 3, 4, 5, 6, 7, 8, 9$ . The number of ways in which this can be done is

$9!$

$2 (7!)$

$4 (7!)$