find the sum of all four digits numbers that can be formed using the digits 0,1,2,3. no digits can be repeated.

• sum of all no. formed=(n-1)!*(sum of n digits)*(111..n times)
  =3!*(6)*(1111)
  =39996
• 9 years agoHelpfull: Yes(20) No(1)
• Sum of all no. formed that are distinct = (n-1)! * (sum of n digits) * (111.. n times)
  = 3! * (0+1+2+3) * (1111)
  = 39,996
  The above includes when 0 is in thousands place.
  
  Hence we need to subtract that value, which means we need to find the sum of all the numbers formed using 1,2,3
  Using the above formula we get
  =2! * (3+2+1) * (111)
  =1332
  
  Hence 39,996-1,332 = 38,664
  
  38,664 is the answer
• 9 years agoHelpfull: Yes(10) No(0)
• 3012
  3021
  3120
  3102
  3201
  3210
  Sum is 18666
  2310
  2301
  2031
  2013
  2103
  2130
  Sum is 12888
  1320
  1302
  1023
  1032
  1203
  1230
  Sum is 7110
  Total sum = 18666+12888+7110=38664
• 9 years agoHelpfull: Yes(8) No(1)
• 1+2+3+0=6
  (4-1)!
  1111
  6*3!*1111=39996
  
• 9 years agoHelpfull: Yes(4) No(2)
• 3!*1000(3+2+1) + 2*2*1*100(3+2+1) + 2*2*1*10(3+2+1) + 2*2*1*(3+2+1) { + 3!*100(0) + 3!*10(0) + 3!(0)} = 38664
  PS: the term inside the braces =0 . Added just for better understanding & explanation
• 9 years agoHelpfull: Yes(4) No(0)
• 3*3*2*1=18
• 9 years agoHelpfull: Yes(1) No(1)
• (0+1+2+3)(1111)(6)-(0+1+2+3)(111)(2)
  =38664
• 9 years agoHelpfull: Yes(1) No(1)
• total sum=4(3)(1111)(6)=79992
• 9 years agoHelpfull: Yes(0) No(2)
• Munish yadav..
  is there any shortcut method.
• 9 years agoHelpfull: Yes(0) No(0)