Q. How many 4 digit number are their which have exactly 3 distinct digits

• 0,1,2,3,4,5,6,7,8,9,
  not started with 0
  in first position= we can arrange in 9 way
  2nd position = arrange in 9 way
  3rd position= arrange in 8 way
  4th position to make only 3 digit different we have only three number left to arrange
  9*9*8*3=1944
  is it right?
  
• 10 years agoHelpfull: Yes(37) No(13)
• answer is 3888
  
  9*9*8*6=3888
  
  all of u are wrong
• 10 years agoHelpfull: Yes(17) No(7)
• 0,1,2,3,4,5,6,7,8,9,
  not started with 0
  in first position= we can arrange in 9 way
  2nd position = arrange in 9 way
  3rd position= arrange in 8 way
  question condition is exactly 3 distinct sigits so
  4th position= arrange in 7 way
  9*9*8*7=4536
  is it ri8 ?
  
• 10 years agoHelpfull: Yes(17) No(3)
• 1st position in 9 ways
  2nd postion z also in 9 ways
  3rd postion z in 8 ways
  but for 4th position in 10 ways
  (he gave exactly 3 distnct digits 4th digit may r may not be same dat means in 0 ways we can select the 4th digit)
  ans=9*9*8*10
• 10 years agoHelpfull: Yes(13) No(3)
• 1,2,3,4,5,6,7,8,9,0
  not started with 0
  in first position= 9!
  2nd position =9!
  3rd position=8!
  4th position=3
  a number must be chosen from used digits, to make only three distinct digits
  9!*9!*8!*7!
  
• 10 years agoHelpfull: Yes(2) No(7)
• 10c3= 120 - of ways to choose 3 distinct digits out of 10;
  3c1= 3 - of ways to choose which digit will be represented twice;
  4!/2!= 12 - of permutation of digits aabc;
  
  10c3*3c1*4!/2!=4320
• 10 years agoHelpfull: Yes(2) No(2)
• Case I:
  The repeated digit is the unit's digit.
  So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 4th digit (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.
  Thus, in all we have:
  9x9x8x3
  
  Case II:
  The repeated digit is the ten's digit.
  So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 3rd digit (ten's digit) can be either equal to the 1st or 2nd digit.
  Thus, in all we have:
  9x9x2x8
  
  Case III:
  The repeated digit is the hundred's digit.
  So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 2nd digit (hundred's digit) is equal to the 1st digit.
  Thus, in all we have:
  9x1x9x8
  
  In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = 3888
  
  Hope this helps.
• 10 years agoHelpfull: Yes(2) No(1)
• Sorry @Jai...u r right... I hv msd this part that 3 distinct digit there are ....
• 10 years agoHelpfull: Yes(1) No(4)
• i got call for banglore venue 25 jan may i go to for delhi venue 25 jan
• 10 years agoHelpfull: Yes(1) No(15)
• 3888 is ans
  
• 10 years agoHelpfull: Yes(1) No(0)
• first combinarion of 3 numbers,then make permutation of 3p3 and is equal to 6.
  then make 3 set of 6 numbers.to each number of each set add one of three given numbers with different arrangements.that makes totally 24+24+24=72.
  suppose numbers are 1,2,3.number should be of these numbers.meaning no 4,5,6,....9 in numbers.one number will repeat in each number.
• 10 years agoHelpfull: Yes(0) No(4)
• 1,2,3,4,5,6,7,8,9,0
  not started with 0
  in first position= 9!
  2nd position =9!
  3rd position=8!
  4th position=7!
  9!*9!*8!*7!
  is it right?
• 10 years agoHelpfull: Yes(0) No(12)
• ANSHUL KUMAR
  
  can u pls explain it with explanation.......
• 10 years agoHelpfull: Yes(0) No(2)
• ans is 6480=9*10*9*8
• 10 years agoHelpfull: Yes(0) No(2)
• ans is=8*9*9*9=5832
• 6 years agoHelpfull: Yes(0) No(1)