How many numbers can be formed which is greater then 6*10^6 using 5,5,6,6,9,9,0 ?

• (4*6*5*4*3*2*1)/(2!*2!*2!)=360
• 9 years agoHelpfull: Yes(39) No(1)
• at unit place if we take 6 then no of solution= 6!/(2!*2!) =180
  at unit place if we take 9 then no of solution = 6/(2!*2!) =180
  hence total is 360.
• 9 years agoHelpfull: Yes(20) No(3)
• (6!+6!)/(2!*2!)=180
• 9 years agoHelpfull: Yes(9) No(5)
• here seven place available -- -- -- -- -- -- -- -- but for 1st place choose only four value 6 , 6, 9 ,9 fulfill the condition 6*10^6=6000000 and for 2nd 6 option , 3rd 5 option ,4th 4 option ,5th 3 option,6th 2 option , 7th 1 option so answer is --> 360
• 9 years agoHelpfull: Yes(9) No(0)
• for a number greater than 6*10^6=6000000 from given digits it shall start from 6 or 9
  for 6,the other 6 digits may be arranged as 6*5*4*3*2*1=720(if repetition is not allowed)
  for 9,again the same pattern follows=720
  total number of numbers=720+720=1440
• 9 years agoHelpfull: Yes(3) No(2)
• The ans. will be 360.
• 9 years agoHelpfull: Yes(2) No(2)
• i think answer should be 180...
• 9 years agoHelpfull: Yes(1) No(1)
• There was no option like 360 or 180.
• 9 years agoHelpfull: Yes(0) No(1)
• plz tell me exact answer of dis ques....
  
• 9 years agoHelpfull: Yes(0) No(0)
• 2880 i think
  4* 6! = 4* 720 = 2880
• 9 years agoHelpfull: Yes(0) No(1)
• avinash roy is correct...bt....in above fact should be there...
• 9 years agoHelpfull: Yes(0) No(0)
• The right answer is 1800.
  how ? :
  no. greater than 600000 can be of 6 digit or 7 digit.
  for 6 digit no :
  1st digit must be 6 or 9 by ways = 2
  2nd digit can be 5,6,9 or 0 by ways = 6
  3rd digit can be 5,6,9 or 0 by ways = 5
  4th digit can be 5,6,9 or 0 by ways = 4
  5th digit can be 5,6,9 or 0 by ways = 3
  6th digit can be 5,6,9 or 0 by ways = 2
  but if 6 digit no. leaves 0 then no. of ways to form digit = (2*6*5*4*3*2)/(2!*2!*2!)
  but if 6 digit no. leaves single 5 then no. of ways to form digit = (2*6*5*4*3*2)/(2!*2!)
  but if 6 digit no. leaves single 6 then no. of ways to form digit = (2*6*5*4*3*2)/(2!*2!)
  but if 6 digit no. leaves single 9 then no. of ways to form digit = (2*6*5*4*3*2)/(2!*2!)
  thus total no. of way to form 6 digit no greater than 600000 using 5,5,6,6,9,9,0 is :
  (2*6*5*4*3*2)/(2!*2!) + (2*6*5*4*3*2)/(2!*2!) + (2*6*5*4*3*2)/(2!*2!) + (2*6*5*4*3*2)/(2!*2!*2!)
  = 6!* (4+4+4+2) / (2!*2!*2!)
  = 720 * 14 / 8
  = 90 * 14 = 1260
  for 7 digit no.:
  no of ways = 6 * 6! / (2!*2!*2!) = 6 * 720 / 8 = 6 * 90 = 540
  hence
  total no. form using 5,5,6,6,9,9,0 which are greater than 600000 = 540 + 1260 = 1800.
• 9 years agoHelpfull: Yes(0) No(2)