Consider set of 4 digits numbers all of whose digits are distinct (The number cant start with 0 but may have 0 among other digits) .For example the set will contain 1492 & 1234 but not 1231.If we list all the numbers in the ascending order how many number will come before 1239?
a)71 b)70 c)69 d)68

• And is 69
  Like when it is like 10**-> in star places I.e in tens place u cant take 0,1...so there are 8 no's posibilits ..and in unit plc u cant 1,0 and the digit alrdy taken in tens plc...so total no's can be psbl 7*8=56
  Now u cant take any no's btwn 1100-1199.,,,now when 12**
  1200 - 1210-> 7nos can be psbl
  1210 - 1219 -> 0 psbl
  1220-1229-> 0 psbl
  1230-1238-> 6 psbl
  So total 69
• 8 years agoHelpfull: Yes(17) No(1)
• answer is 70
  Let's see how:
  first digit is always 1therefore 1_ _ _. , now lets fill in the next 3 digits
  second digit can be only 0 or 2 as 1 is already a first digit, lets first place 0 and find out how many nos. can be formed 1 0 _ _==> 8 digits can be placed at 3rd position and 7 digits at 4th position
  i.e. 8*7=56 combinations are possible for 1 0 __ __
  now lets place 2 at second position 1 2 __ __==> as the no must be less than 1239 therefore
  seven combinations are possible for 1 2 0 __ and seven are possible for 1 2 3 __
  therefore total possible nos. less than 1239 are =56+7+7=70
• 8 years agoHelpfull: Yes(5) No(9)
• answer is 69
  first digit is always 1therefore 1_ _ _. , now lets fill in the next 3 digits
  second digit can be only 0 or 2 as 1 is already a first digit, lets first place 0 and find out how many nos. can be formed 1 0 _ _==> 8 digits can be placed at 3rd position and 7 digits at 4th position
  i.e. 8*7=56 combinations are possible for 1 0 __ __
  now lets place 2 at second position 1 2 __ __==> as the no must be less than 1239 therefore
  seven combinations are possible for 1 2 0 __ and six combinations are possible for 1 2 3 __
  therefore total possible nos. less than 1239 are =56+7+69=69
• 8 years agoHelpfull: Yes(5) No(0)
• i think the ans is 69
• 8 years agoHelpfull: Yes(3) No(0)
• 8*7 is correct and after that 1203,04,05,06,07,08,09 then 1234,5,6,7,8 so 56+7+5=68...all should be distinct and no nos shld b repeated..all u guys hav calculated wrngly..
• 8 years agoHelpfull: Yes(3) No(0)
• Consider a 4 digit number __ __ __ __
  Its Thousand place can be filled with 1 digit only.i.e 1
  Because that 4 digit number must be less 1239 and we can not 0 at the thousand place.
  Again at hundred place we can fill it with 3 digits 0,1 & 2 we can not take other values because the number will exceed 1239.
  In the same manner ten place can be filled with 4 digits 0,1,2,3
  And now unit place can be filled with 6 digits in the same manner because all the digits are distinct.
  So total numbers 1*3*4*6=72
  it include 1239 also and the question asked numbers before 1239 so,
  72-1=71 is the correct answer.
• 8 years agoHelpfull: Yes(2) No(9)
• @ GAURAV u cannot take 7 possibles after 123_.. The possibles are 0,4,5,6,7,8... AS it is asked the terms before 1239 not wd 1239..... Hence The answer is 69.
• 8 years agoHelpfull: Yes(1) No(0)
• think then judge the answer
  
  10__ __ can be filled by 8*7 ways=56
  
  11___ __ can not take
  
  12__ __ can be filled by (0,1,2,3) & (0,1,2,3,4,5,6,7,8) since we have already taken (1,2) so 100th place can be filled by 2 ways(0,3) & 1000th place can be filled by((0,1,2,3,4,5,6,7,8)-(3 numbers))=6 ways
  so,2*6=12 ways
  
  total ways=56+12=68 ways
  
  
• 8 years agoHelpfull: Yes(1) No(1)
• i missed out a number 1230 you can have 69 possibilities then...sorry for wrong calculation
• 8 years agoHelpfull: Yes(1) No(0)
• 69 is the ans
• 8 years agoHelpfull: Yes(1) No(0)
• 69 is the answer
• 8 years agoHelpfull: Yes(0) No(1)