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  • Stuck here, help me understand: An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is

 An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is

  • Option 1)

    72(7!)

  • Option 2)

    18(7!)

  • Option 3)

    40(7!)

  • Option 4)

    36(7!)

 

Answers (1)

best_answer

As we have learned

The Number of ways of Arrangement of objects -

The number of ways of n different objects taken all at a time =\ ^{n}p_{n}=n!

- wherein

Where 0! = 1

 

 0+1+2+......+9= 45  ,which is divisible by 9 now we have to select 8 numbers , which means we have to reject 2 nos while keeping the sum divisible  by 9 ...(0,9), (1,8) (2,7), (3,6) ,(4,5) can be rejected 

1) if (0,9 ) is rejected , we can form numbers is 8 ! ways 

2) If other 4 sets are rejected , we can form numbers in 7 (7!)  we can't be at first place 

so no. of ways = 4* 7 (7!)

Adding (1) and (2) .....(8)! + 7(7!)= 36 (7!)


Option 1)

72(7!)

Option 2)

18(7!)

Option 3)

40(7!)

Option 4)

36(7!)

Posted by

gaurav

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