A seven digit number is formed by using the digits 3, 3, 4, 5, 5, 8 & 9. Probability that the number formed is great than 57,26,000 is?
OPTION 1) 3/8
2) 1/5
3) 1/3
4) 2/9
5) 12/35
6) 16/35
7) 8/35
8) 8/21
9) 2/5

• only four condition...
  58........=====> 3 is repeated and first two no is fixed..then..5!/2....
  
  59.........=====>3 is repeated and first two no is fixed..then..5!/2....
  
  8.......=> in this case both 3 and 5 is repeated..so..= 6!/(2*2)
  
  9......===>.in this case both 3 and 5 is repeated..so..= 6!/(2*2)
  lower part is 7!/(2*2),,because of repeatation....
  so result==>uper part sum..==5!/2 + 5!/2 + 6!/(2*2) + 6!/(2*2)
  ===> 5!*4
  then result===5!*4/(7!/2*2)==5!*4*4/7!==4*4/(7*6)==8/21 Ans..
• 10 years agoHelpfull: Yes(23) No(2)
• total sample space is 7!/(2!*2!)=1260
  total number of numbers formed by 3,3,4,5,5,8,9 which are greater than 5726000 are
  when first digit is 5 second digit should be 8 or 9
  so 58*****=5!/2!=60
  and 59*****=5!/2!=60
  or first digit should be 8 or 9
  so 8******=6!/(2!*2!)=180
  and 9******=6!/(2!*2!)=180
  so required probability = (60+60+180+180)/1260=480/1260=8/21
• 10 years agoHelpfull: Yes(7) No(1)
• ans is 1/3
  total no.of outcome=7!=7*6*5*4*3*2*1
  case 1 :5 8 _ _ _ _ _
  or
  case 2 :5 9 _ _ _ _ _
  or
  case 3:8 _ _ _ _ _ _
  or
  case 4:9 _ _ _ _ _ _
  these 4 cases are favourable no. of outcome
  there fore required probability=5!/7!+5!/7!+6!/7!+6!/7!=1/3 ans
• 10 years agoHelpfull: Yes(5) No(11)
• 8/21
  soln:
  - - - - - - - the no of ways we can the dashes with these numbers (3, 3, 4, 5, 5, 8 & 9.) is 7!/(2!*2!)=1260
  The number of numericals greater than 57,26,000 is 58-----(5!/2!) or 59-----(5!/2!) or 8------(6!/(2!*2!)) or 9------(6!/(2!*2!))
  sum of all these is 480
  probability = 480/1260= 8/21
• 10 years agoHelpfull: Yes(3) No(0)
• Absolutely Correct Solution......
  Total no. of numbers formed by using the given digits 3,3,4,5,5,8,9 = (7*6*5*4*3*2*1)/(2*1)*(2*1) = 1260
  (Denominator part is due to the repetition of 3 and 5)
  Total no. of numbers formed by using the given digits 3,3,4,5,5,8,9 which are greater than 57,26,000 are of two types
  Total no. of numbers started with 5 = (2*2*5*4*3*2*1)/ (2*1)*(2*1) = 120
  Total no. of numbers started with 8&9 = (2*6*5*4*3*2*1)/ (2*1)*(2*1) = 360
  Hence required probability = (120 + 360)/ 1260 = 480/1260 = 8/21
• 10 years agoHelpfull: Yes(3) No(0)
• please provide proper solution
• 10 years agoHelpfull: Yes(0) No(0)
• according to me . no of favourable outcomes would be 5!+5! cz the no greter than 5726000 would hav either 58 in thebegining or 59 in the begining
  
  total no of outcomes would be 7!
  
  probability =2*5!/ 7!
  =1/21
• 10 years agoHelpfull: Yes(0) No(4)
• total no. of 7 digit no.s = 7!/(2!*2!)=1260
  case 1 : 5,8 _ _ _ _ _=240/4=60
  case 2 : 5,9_ _ _ _ _ =240/4=60
  case 3 : 8,9,_ _ _ _ _ =120/4=30
  case 4: 9,8,_ _ _ _ _ =120/4=30
  p= (60+60+30+30)/1260=1/7
• 10 years agoHelpfull: Yes(0) No(1)
• There can be another case.. 57-------.
  THEN (7*6*5*4*3/2*2)/(7!/2*2)+8/21
• 10 years agoHelpfull: Yes(0) No(0)
• - - - - - - - the no of ways we can the dashes with these numbers (3, 3, 4, 5, 5, 8 & 9.) is 7!/(2!*2!)=1260
  The number of numericals greater than 57,26,000 is 58-----(5!/2!) or 59-----(5!/2!) or 8------(6!/(2!*2!)) or 9------(6!/(2!*2!))
  sum of all these is 480
  probability = 480/1260= 8/21
• 10 years agoHelpfull: Yes(0) No(0)