Given digits 2,2,3,3,4,4,4,4 how many distinct 4 digit numbers greater than 3000 can be formed?

• As our four digit numbers have to be greater than 3000, so we have only two options -- the numbers will start with 3 or 4.
  
  CASE 1: (Format of the no. is 3 * * *)
  Now we have the digits 2, 2, 3, 3, 4, 4, 4, 4
  Now if we consider that there are one 2 and one 3 more, then the blank spaces could be filled up in 3X3X3 ways, i.e. in 27 ways.
  But we have two options less, as we cannot form 3222 and 3333.
  So in this case, total no. of numbers is (27 - 2) = 25.
  
  CASE 2:(Format of the no. is 4 * * *)
  Now we have the digits 2, 2, 3, 3, 3, 4, 4, 4
  Now if we consider that there is one 2 more, then the blank spaces could be filled up in 3X3X3 ways, i.e. in 27 ways.
  But we have one options less, as we cannot form 4222.
  So in this case, total no. of numbers is (27 - 1) = 26.
  
  Here from the above two cases, we see that there are (25 + 26) = 51such numbers.
  
• 12 years agoHelpfull: Yes(25) No(25)
• 44 such numbers can be formed.
  
  Possible 1st digit is 3 and 4.
  
  Case 1: First digit is 3. (3 * * *)
  Remaining digits to use = 2,2,3,4,4,4,4 (In this have to chose 3digits)
  
  -> 3 alike digits = 1 case(all 4's)
  -> 2 alike 1 different digits = 2*2*3!/2! = 12 cases
  -> 3 different digits = 1*3! = 6 cases
  -> Total such numbers = (1+12+6) = 19 numbers.
  
  
  Case 2: First digit is 4. (4 * * *)
  Remaining digits must come from the list: 2, 2, 3, 3, 4, 4, 4
  
  Therefore we may choose any 3-digit sequence except 222 and 333 for the rest of the digits. This shows there are
  
  3*3*3 - 2 = 25 such numbers.
  
  So, total numbers = 19 + 25 = 44 such numbers.
• 12 years agoHelpfull: Yes(11) No(6)
• @ Chaitanya,
  
  Pls check.
  consider first case, we have only two 3s.
  so 3233,3433,3332,3334 3323, etc not possible.
  in second case 4333 is also not possible.
• 12 years agoHelpfull: Yes(6) No(7)
• For a number to be greater than
  3000, 3 or 4 should occupy
  thousandth position.
  When 4 is in thousandth position,
  Required number of ways =
  1*3*3*3-2 = 25 (Remember only
  two 2s and two 3s are there.
  Hence 222 and 333 are not
  possible. So we subtract 2)
  When 3 is in thousandth position,
  to fill the remaining three
  positions (hundredth, tenth and
  unit)
  When 3 digits are same (can
  happen only when 4's alone
  used), num of ways = 1
  When 2 digits are same and 1
  digit is unique, the numbers that
  can be considered are 2,2,3,4,4.
  Num of ways = 1*3*2*2 = 12
  When all 3 digits are unique, num
  of ways = 3! = 6
  Total num of ways = 1+12+6+25
  = 44
• 11 years agoHelpfull: Yes(1) No(4)
• the numbers can start with 3 or 4
  case 1: 1st digit is 3_ _ _
  the possibilities that can occur are 3444(1),3442(6!/2!),3443(3),3422(3),3423(6)
  the brackets indicates the possible ways of arranging them.suppose consider 3442 as 1st digit is fixed, 4,4,2 can arranged in 6!/2! ways i.e 3 ways.
  so the numbers that can be formed with first digit 3 is16
  case 2:1st digit is 4: 4_ _ _
  the possibilities are 4444(1),4443(3),4442(3)4433(3),4432(6),4422(3),4322(3),4332(3).so the numbers with 1st digit 4 are 25
  so total 25+16=41 numbers
• 12 years agoHelpfull: Yes(0) No(5)
• 36
  8c4-8c2-2-4=36
• 12 years agoHelpfull: Yes(0) No(3)