How many 8 digit numbers can be formed with 0- to 9 if repetition is not allowed?which is divisible by 9?

• 181440 eight digits numbers can be formed.
  
  Long Explaination :
  
  Rule For the number to be divisible by 9, the digits must add up to a multiple of 9.
  
  Here We have 10 digits = 0,1,2,3,4,5,6,7,8,9
  
  0+1+2+3+4+5+6+7+8+9 = 45(which is divisible by 9).
  
  To form a 8 digits numbers, we want to remove 2 digits from those 10digits. And that two digits must be add up to be a multiple of 9.
  
  Therefore, possible 2 digits we omit which multiple by 9 are,
  (0,9) (1,8) (2,7) (3,6) (4,5) =>(5 Sets are there)
  
  When we omit (0,9) pair, possible ways to arrange a number = 8!
  
  When we use (0,9) pair means, possible ways to arrange a 8digit number = 7(7!)
  
  (Because, 0 is not placed in first position but it can placed in rest of 7 digits.)
  
  That means, we have ONE choice(omitting 0,9) to placed 8! numbers.
  And, we have FOUR choices for 7(7!) numbers.
  
  Finally. we adding them = 8!+(4×7)(7!)
  =8(7!)+28(7!)
  =36(7!)
  =36(7*6*5*4*3*2*1)
  =36(5040)
  =181440
  
  So, 181440 eight digits numbers can be formed bu using 0-9 digits without repetition.
  
  
• 12 years agoHelpfull: Yes(12) No(1)
• I guess 181440 eight digit numbers can be formed which is divisible by 9 by using digits from 0-9(repetition is not allowed)
• 12 years agoHelpfull: Yes(5) No(1)