how may number are there greater than 6*10^6 by using digits 5,6,9,0 . repetition is not allowed?
a-180
b-520
c- not remember

• question is wrong ,
  Correct question is that asked in elitmus ..
  how may number are there greater than 6*10^6 by using digits 5,5,6,6,9,9,0 when repetition is not allowed?
• 8 years agoHelpfull: Yes(34) No(5)
• 7 places - _ _ _ _ _ _ _
  using permutations rule
  1st place - (6,6,9,9) - 4!
  similarly,
  in order of places- 4 6 5 4 3 2 1
  now, since (2,6,9) have commons -
  (4*6*5*4*3*2*1)/(2!*2!*2!) = 360
• 8 years agoHelpfull: Yes(33) No(7)
• Numbers should be greater than 6*10^6
  
  It means numbers should start with 6 and 9(from the given digits only this two)
  
  Starting with 6
  6 _ _ _ _ _ _
  above 5 places can be filled with remaining 6 digits in 6!/(2!*2!) =>180
  
  Starting with 9
  9 _ _ _ _ _ _
  above 5 places can be filled with remaining 6 digits in 6!/(2!*2!) =>180
  
  Total =180+180 =360
• 8 years agoHelpfull: Yes(30) No(4)
• and its answer is 360.
  (4*6!)+(2!*2!*2!*2!)
• 8 years agoHelpfull: Yes(9) No(20)
• 360..iam i right!
• 8 years agoHelpfull: Yes(2) No(7)
• since there are 7 places and the no has to be greater than 6*10^6
  and the no are 5,5,6,6,9,9,0
  _ _ _ _ _ _ _=2*6*5*4*3*2*1=180
• 8 years agoHelpfull: Yes(2) No(3)
• this is not possible as it has to bea seven digit no and not possible with 4 digits
  
• 8 years agoHelpfull: Yes(1) No(8)
• 7 places - _ _ _ _ _ _ _
  1st place - (6 or 9) = 2
  similarly,
  second place there is (5,6,9,0) = 4
  same as other places
  
  2*4*4*4*4*4*4 = 8192 answer
  
  i think some part of question is missing that is why answer is not in options
  
  
• 8 years agoHelpfull: Yes(1) No(6)
• @saurabh rathour can u explain the approach plz ?
• 7 years agoHelpfull: Yes(1) No(1)
• bakwas site hai , genuine question ko change karke puchha gaya hai, please don.t use this site anymore
• 7 years agoHelpfull: Yes(1) No(3)
• 520 bt how i dnt know
• 8 years agoHelpfull: Yes(0) No(10)
• since 7 digits are given in the question so any seven digit number formed from the combination of these digits (except the ones starting from zero ) would be greater than 6*10^6.
  so the correct method is total of seven digits no. + six digits no. (starting from 0) greater than 6*10^6
  i.e. (6*6!)/(2!*2!*2!) + (4*5!)/(2!*2!*2!*)
• 8 years agoHelpfull: Yes(0) No(0)
• use this formula n!/(n-r)! n:6 r:4 have fun
• 7 years agoHelpfull: Yes(0) No(2)
• In the question it is given that digits should not be repeted. So how can we use Two five's,two nine's.?
• 7 years agoHelpfull: Yes(0) No(0)