The no of digits greater than 6 * 10^6 that can be formed by 0,5,5,6,6,9,9 is:
OPTIONS
a)320
b)540
c)720
d)dont remember

This question was asked on 22nd feb ph test

• I think the question is how many 7 digit nos can be formed using digits 0,5,5,6,6,9,9 which are greater than 6(10^6) i.e., 6000000.
  Let the 7-digit nos be _ _ _ _ _ _ _
  To get a number greater than 6(10^6) , we can fill the first blank with either 6 or 9.
  Case 1:
  The first digit is filled with 6 and the remaining digits can be filled in 6!/(2! *2! ) ways {because it is a permutation of 6 objects out of which 2 objects are similar (5,5) ,another 2 objects are similar(9,9) and the remaining are different (0,6) }
  Therefore no of such 7 digit nos which start with 6 are 720/4 =180
  
  case 2:
  similarly we can have the first digit as 9 and remaining 6 digits are permuted.
  In that case no of such nos is also 6!/(2! 2!) = 180
  
  So total nos that can be formed which are greater than 6(10^6) = 180+180 = 360
  Answer is 360.
• 9 years agoHelpfull: Yes(33) No(0)
• 6_ _ _ _ _ _ here empty can be use 055699 that can be arranged 6!/2! 2! =180
  9_ _ _ _ _ _ can also used 055669 that can be arranged 6!/2! 2!=180
  so total number are 180+180=360
  
• 9 years agoHelpfull: Yes(21) No(0)
• case 1: when left most side will fill with 6 then
  2*6*5*4*3*2*1/2!*2!*2! = 180
  case 2: when left most side will fill with 9 then
  2*6*5*4*3*2*1/2!*2!*2! = 180
  so, total no formed = 180+180 = 360
  
• 9 years agoHelpfull: Yes(10) No(1)
• (7! / 2!*2!*2!) - (6! / 2!*2!) - (6! / 2!*2!*2!)
  =630-180-90
  =360
• 9 years agoHelpfull: Yes(2) No(1)
• for the number to be greater than 6*10^6 i.e 6000000, the left most number should be either 6 or 9.
  like for example: 6055699 or 9055669
  consider the first set of number which begins with 6. In that we have the left most number fixed and the rest 6 digits that can be varied.
  so its 6*5*4*3*2*1 = 720.
  but in that numbers we can see that 5 is repeated twice and 9 is repeated twice, hence we need to divide by 2! twice: the actual solution will be (6*5*4*3*2*1)/(2! *2!) which is equal to 180.
  this 180 is the possibilities with starting digit as 6.
  
  the same has to be done for 9 as well..
  even in that numbers the starting number is fixed and the rest 6 digits are variable.
  again here 5 is repeated twice , and 6 is repeated twice. hence the solution will be
  (6*5*4*3*2*1)/ (2! * 2!) = 180.
  
  total we have 180+180=360
  hence the answer is 360
• 9 years agoHelpfull: Yes(2) No(0)
• 4*6!/2!*2!*2!
• 9 years agoHelpfull: Yes(0) No(0)
• (4*6*5*4*3*2*1)/(2*2*2)=360
  ans (d)
• 9 years agoHelpfull: Yes(0) No(0)
• total available digits = 7 (0,5,5,6,6,9,9)
  __ __ __ __ __ __ __
  places : 1st 2nd 3rd 4th 5th 6th 7th
  
  for 1st position possible numbers are (6,6,9,9) so 4
  because of repetition 4/(2! * 2!)
  
  for places 2nd to 7th
  6*5*4*2*1
  because of repetition remaining number i.e. (5,5) = (6*5*4*2*1)/2!
  
  altogether 4* (6*5*4*2*1) / (2! * 2! * 2!) = 360
• 4 years agoHelpfull: Yes(0) No(0)