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

• From these four digits '0' cannot be at first place to form four digit number so we have 1, 2, or 3 can be at thousand's place keeping one digit at thousand place can have 6 different four digit number so that means 1000*6 + 2000* 6 + 3000* 6 = 36000.
  Now come to hundred's place we can have all four digit at this place , '0' will not contribute any thing to sum so in all four digits number 1,2,3 will be at 4 times on same place so that means 100*4 + 200* 4 + 300 * 4= 2400.
  Now come to tenth's place ...same as hundred so 10*4 + 20*4 + 30*4= 240.
  same for unit place 1*4 + 2*4 + 3*4= 24 in last if you add whole the total sum of all possible four digit number formed is 38664
• 9 years agoHelpfull: Yes(26) No(3)
• (n-1)!*1111*(sum)
  (4-1)!*1111*6
  39996
• 9 years agoHelpfull: Yes(25) No(4)
• three cases-
  when 1 in thousands place=
  1 0 2 3
  1 0 3 2
  1 2 0 3
  1 2 3 0
  1 3 0 2
  1 3 2 0
  similarly for 2 and 3 in thousands place- 6-6 cases each
  total num=6+6+6=18
  
  if we see, in units, tens and hundreds place-
  4 times 1=4
  4 times 2=8
  4 times 3=12
  6 times 0=0
  so sum at units,tens and hundreds place=4+8+12=24, 24, 24
  
  thousandth place-
  6 times 1=6
  6 times 2=12
  6 times 3=18
  sum at thousnds=6+12+18=36
  
  now sum = 36*1000 + 24*100 + 24*10 + 24*1 = 38664
  
• 9 years agoHelpfull: Yes(20) No(3)
• Sum of all the r-digits formed by taking the given n-digits = (sum of all n-digits ).n-1Pr-1*(1111.....r times)
  So here n=4 and r=4
• 9 years agoHelpfull: Yes(3) No(5)
• total non repeated numbers:3*3*2*1=18
  sum of all numbers will be:(1000)*6*18+(100)*6*18+(10)*6*18+(1)*6*18
  =108*(1111)
  =119988
• 9 years agoHelpfull: Yes(2) No(1)
• since 0 is present the formula becomes
  (n-1)!*(1111)(sum of digits)-(n-2)!*1111(Sum of digits)
  1111*7*(6-2)=31108
• 9 years agoHelpfull: Yes(2) No(0)
• Any way and = 0+1+2+3=6
• 9 years agoHelpfull: Yes(1) No(2)
• as there is zero in the given nos the formula is ((n-1)!*(1111)*(0+1+2+3))-((n-2)!*(1111)*(0+1+2+3)=26664
• 9 years agoHelpfull: Yes(1) No(0)
• 4! i.e equals to 4x3x2x1=24
• 9 years agoHelpfull: Yes(1) No(2)
• For thousadth position=1000(1+2+3+0)*3*4*4*4
  =1000(6)*192
  For hundredth position sum=100(6)192
  For Tenth position sum=10(6)*192
  Unit position sum=1(6)8192
  overall sum=(6)192(1000+100+10+1) =1279872
• 9 years agoHelpfull: Yes(0) No(7)
• lowest 4 didgit no :1000
  highest 4 digit no :3333
  so sum on all numbers from 1000-3333 = (3333*3334)/2-(999*1000)/2
  here we are finding sum of numbers from 1-3333 and then removing from 1-999
  Formula = (n*n+1)/2 for sum of n numbers
• 9 years agoHelpfull: Yes(0) No(1)
• no 0 is allowed in thousend position
  so ans=>38664
• 9 years agoHelpfull: Yes(0) No(0)
• answer is 56664
• 9 years agoHelpfull: Yes(0) No(0)
• no digit repeated so,at every place 4 numbers in 4 ways...
  so 4*4*4*4=256 ...i think
• 9 years agoHelpfull: Yes(0) No(3)
• 38664... as there would be 18 different arrangements with taking number 1 , 2,3 in thousands place.... adding these all no would result in no 38664
• 9 years agoHelpfull: Yes(0) No(0)
• plzzz tell how many pages i have to solve so that iam able to solve tcs one....
  in set of 30 quesytion
• 9 years agoHelpfull: Yes(0) No(0)