How many three digit numbers abc are formed where two of the three digits are same.
a.dnt rember b.243 c.251 d.dnt rember

• total no 3 digit = 9*10*10 =900
  total no no 3 digit without repeation are 9*8*7=648
  so atleast 2 digit same means atleast 0ne or 2 digit r same so 900-648=252
• 9 years agoHelpfull: Yes(23) No(36)
• Ans is 243
  Expln:
  total three digits number with repeating is 9*10*10=900
  total three digits number where three digits are same=9
  total three digits number where digits are not same 9*9*8=648
  
  so number of three digit number where two digits are same is 900-(648+9)=243
• 8 years agoHelpfull: Yes(23) No(3)
• I thing correct ANS is 243
  Because between 100 to 200 three digit no are 27
  27*9=243
• 9 years agoHelpfull: Yes(13) No(8)
• Hello listen
  
  digits are 0,1,2,3,4,5,6,7,8,9
  so no. of digits are 10
  first all possible case =>9(0 excluded)*10*10=900
  second no repetition allowed =>9*9*8=648
  third all digits are same =>9(111,222,333,444,555,666,777,888,999)
  final answer 900-648-9=243 ;
  
• 8 years agoHelpfull: Yes(7) No(2)
• This is a good question.
  If two of the three digits are same.. then our solution reduces to choosing digits to be arranged in only two places becoz anyhow the third digit is going to be repeated.
  Here we have to consider two cases:
  1. zero is present among the three digits
  2. zero is not present.
  Case1:
  arranging 9 digits in two places => 3!*9*8*1/2!
  Case2:
  If a zero is present then it must not be the first digit.
  and here we have again two cases again.
  case1: the repeated number is zero
  case2: the repeated number is any other among 1 to 9.
  Then
  case1: Our solution reduces down to choosing the number for fist place only. Becoz if a zero has to repeat then it must be present in 2nd and 3rd digits only. We just have to choose 1 number among 1-9 => 9 ways.
  case2: here a zero can be present in 2nd or 3rd digit places. Then we have to only choose 1 number that is repeated in two places. This can be done in
  9*1*1*2 ways.
  So total number of ways: 3!*9*8/2! + 9 + 18 = 243
• 9 years agoHelpfull: Yes(6) No(3)
• 243 is ans
• 9 years agoHelpfull: Yes(5) No(4)
• 252 is the right answer
• 9 years agoHelpfull: Yes(5) No(2)
• Sorry option b was 252
• 9 years agoHelpfull: Yes(4) No(2)
• ans is 243,for ex-a number abc having 2 same digits can be of the form-
  11_, 1_1 or _11 now we can fill the blank with any digit from 0-9 except 1 as it wud make 3 digits same
  so we can fill with 9 diff digits in 3 above cases,which makes it 3*9=27
  this applies to digits from 1-9(we can make a 3-digit number with two same digits using 1-9)
  so total=27*9=243
• 9 years agoHelpfull: Yes(3) No(0)
• considering 1 as the repeating digit
  
  11* *can have 0-9 , that is 10 case
  1*1 *can have 0-9 , that is 10 case
  *11 *can have 1-9 , that is 9 case
  
  total case for 1 is 10+10+9=29
  
  for 1,2,3,4,5,6,7,8,9 that is 9 digit
  total case is 29*9=261
  
  which was the option D
• 9 years agoHelpfull: Yes(2) No(5)
• aslam bhai aslam bhai
  Ans is : 243 why ... question said two of three digit not said atleast three digit
  total three digit nos: = 9*10*10=900
  without repetition of 3 digit nos: 9*9*8=648 ( aslam => 9*8*7=504)
  so 3 digit repetition : 9 nos( 111, 222, 333, 444, 555, 666, 777, 888, 999)
  so i want only 2 digit repetition nos: 900-(648+9)= 243
  *** why m4maths give best solution and also calculation missing... because elitmus always some tricky change in question.......
• 7 years agoHelpfull: Yes(2) No(0)
• 10*9+2*9*9=252
• 9 years agoHelpfull: Yes(0) No(1)
• no. of 3 digit no.s with two of the digits same = total no. of 3 digit no.s possible - (no. of 3 digit no.s with no digits same + no. of 3 digit no.s with all the digits same...i.e, 111..222...999 )
  = (9*10*10) -( (9*8*7) + 9 )
  = 900 - (648 + 9 )
  = 900 - 657
  = 243
  hence, 243 is the correct answer.
• 8 years agoHelpfull: Yes(0) No(0)
• 244
  numbers can be 11x,22x,33x...99x(9 ways) where x can take 9 values. Also a number can be arranged in 3!/2! ways
  So total number of digits= 9*9*(3!/2!)=243.
  Now 100 is also there. So total number of digits=244
• 8 years agoHelpfull: Yes(0) No(0)
• three digit no =9*10*10=900
  two digit no are same = 9*9*8 =648
  then three digit no formed = 900-648 =252
• 8 years agoHelpfull: Yes(0) No(1)
• total no 3 digit = 9*10*10 =900
  total no no 3 digit without repeation are 9*8*7=648
  so atleast 2 digit same means atleast 0ne or 2 digit r same so 900-648=252
  now exact two digits same =252-(3 digits are same).
  between 100 to 999 have 9 number that are 3 same digits (111,222,333,444,555,666,777,888,999)
  so reult is 252-9=243
  
  
• 8 years agoHelpfull: Yes(0) No(1)
• take 3 places: _ _ _
  now at 1st place 1-9
  2nd,3 place:0-9
  therefore 2 nos same out of 3,possibility:9x9x1=81
  now the 2 nos can be places at 3 possible places :81*3=243
• 8 years agoHelpfull: Yes(0) No(0)
• we have numbers 0,1,2,3,4,5,6,7,8,9 means total 10 numbers
  Now total three digits numbers can be formed using these 10 digit are
  __ __ __
  9 x 10 x 10 = 900 (As on the first place 0 cant come)
  
  Now total distinct numbers which can be formed using these 10 digits are
  __ __ __
  9 x 9 x 8 = 648
  
  If we subtract Distinct from total we will get all the numbers which have atleast one digit repeative So
  
  900 - 648 = 252 which is our answer
• 7 years agoHelpfull: Yes(0) No(0)
• mistake atleast three digit correct = atleast two digit
• 7 years agoHelpfull: Yes(0) No(0)