HOW MANY 4 DIGITS NUMBER DO NOT CONTAIN 3 OR 6

• 7*8*8*8=3584
  
• 9 years agoHelpfull: Yes(15) No(0)
• suppose the "four digit" number is
  "abcd"
  now
  "d " digit can be 0,1,2,4,5,7,8,9 =8ways
  "c " digit can be 0,1,2,4,5,7,8,9 =8ways
  "b " digit can be 0,1,2,4,5,7,8,9 =8ways
  
  the turnig point is digit "a" can not have {0,3,6}
  because if we use "0" then number becomes 3digit number
  for example
  "0124"
  
  thats why
  
  "a " digit can be 1,2,4,5,7,8,9 =7ways
  
  so
  answer is
  7*8*8*8=3584
  
  
• 9 years agoHelpfull: Yes(9) No(0)
• 2*8*9*9*9=11664
• 9 years agoHelpfull: Yes(2) No(3)
• unit's place can filled in 6 ways except 0,3,6
  and rest 7 ways except 3 or 6
  so, 6*7^3
• 9 years agoHelpfull: Yes(1) No(3)
• its '3 OR 6' not '3 AND 6'
  but most of you have given the answers taking '3 AND 6'....
  am confused ...
  it has to be -->
  no of 4 digit numbers not containing 3 +(OR) no of 4 digit numbers not containing 6
  which is =8*9*9+8*9*9=2*8*9*9
  Ans --> 2*8*9*9
• 9 years agoHelpfull: Yes(1) No(0)
• this can be done in 3 cases-
  1- when 3 is not present then no of ways=8*9*9*9 (since repetition is allowed)
  2- when 6 is not present den no of ways=8*9*9*9 (since repetition is allowed)
  3- when 3 and 6 both are not present den no of ways- 7*8*8*8
  
  total no of ways=sum of all three= 15248 ans.
• 9 years agoHelpfull: Yes(1) No(0)
• 7*8*8*8
  = 3584
• 9 years agoHelpfull: Yes(0) No(1)
• ans is 3854
• 9 years agoHelpfull: Yes(0) No(1)