How many four digit numbers that do not contain the digit 3 or 6

• thousand place can be filled in 7 ways as it cannot start with 0 and it should nt contain 3 and 6,similarly hundreds place is filled by 8 ways(excluding 3 and 6),tens place is filled by 8 ways and unit place is filled by 8 ways.so tot=7*8*8*8=3584
• 11 years agoHelpfull: Yes(51) No(2)
• first no in 4 digits can be filled in 7 ways(1,2,4,5,7,8,9)
  next digit in 8 ways(0,1,2,4,5,7,8,9)
  similarly next two digits in 8 and 8 ways
  so no of ways is 7*8*8*8=3584 ways
  and hence 3584 no are possible
• 11 years agoHelpfull: Yes(9) No(0)
• 7*8*8*8 = 3584
• 11 years agoHelpfull: Yes(3) No(0)
• Hiii evrybody ,there is doubt to solution of this ques;because in a ques condition is digit 3 or 6 should contain not 3 and 6 .that means if we pick up 3 then 6 should not be pick up and vica versa.
  now
  8*9*9*9=5832
  is it right????
• 11 years agoHelpfull: Yes(3) No(7)
• 7*8*8*8 = 3584
• 11 years agoHelpfull: Yes(2) No(0)
• As all positions can have 10 digits each so for thousands place we cant have digits like 0,3,6 so solving according to question thousand place can be filled in 7 ways as it cannot start with 0 and it should not contain 3 and 6,similarly hundreds place is filled by 8 ways(excluding 3 and 6),tens place is filled by 8 ways and unit place is filled by 8 ways.so tot=7*8*8*8=3584
• 11 years agoHelpfull: Yes(1) No(0)
• 7*8*8*8=3584
• 11 years agoHelpfull: Yes(1) No(0)
• no should not contain 3 or 6
  1st case: not having 3 so ways 8*8!
  2nd case: not having 6 so ways 8*8!
  3rd case: neither 3 or 5 so ways 7*7!
  SO total ways 2*8*8!+7*7!
• 11 years agoHelpfull: Yes(1) No(1)
• In the 4 digit no.,the first place can't be filled up by 0,3 and 6.
  so we have 0-9 i.e. 10 digits-3 digits=7 digits to fill the first place
  for 2nd place we can't use 3 and 6
  so we can fill it by 10-2=8 possible ways
  for 3rd and 4th place also,we can fill those in 8 possible ways.
  so total possibility=7*8*8*8=3584
• 11 years agoHelpfull: Yes(1) No(0)
• 3360
  3 or 6 should not occur.
  2*8*7*6*5
• 11 years agoHelpfull: Yes(0) No(8)
• ans is 3584...
  7*8*8*8=3584
• 11 years agoHelpfull: Yes(0) No(0)
• SO I have a doubt with this method you all have used,and y its not applicable for another question!
  
  SO the question is the number of times 2 occurs in 1000 to 9999
  I came across two methods.
  Total 4 digit numbers that we can form is (9*10*10*10)
  Total 4 digit numbers that we can form with out 2 is (8*9*9*9)
  so 4 digit nos with 2 we can form (9*10*10*10)-(8*9*9*9) = 3168
  
  But the general answer given by all were
  Number of 2's at units place(1000 to 9999)=900
  Number of 2's at tenths place(1000 to 9999)=900
  Number of 2's at hundreds place(1000 to 9999)=900
  Number of 2's at thousands place(1000 to 9999)=1000
  
  therefore total number of 2's =(900+900+900+1000)=3700
  
  Both answers are different..and please help me with is the ryt answer
• 11 years agoHelpfull: Yes(0) No(0)
• the first place do not have zero as well as 3 or 6 ..therefore 7 ways(10-3)similarly for other places zero may include ..answer is 7*8*8*8
• 11 years agoHelpfull: Yes(0) No(0)