Question

A seven digit number without repetition and divisible by 9 is to be formed by using 7 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. $2\times 7!$
• C. $4\times 7!$
  
• D. $3\times 8!$

Answer 1

The correct option is C $4\times 7!$


Sum of 7 digits = 9k, $kϵN$
Sum of $1,2,3,\dots \dots 9=45=9\times 5$
So the two remaining left out digits should also have sum of ‘9k’
Possible left out digit pairs = (1, 8), (2, 7), (3, 6), (4, 5)
With each pair, number of 7 digit numbers = 7!
So with all pairs $=4\times 7!$