What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition,

• We know that zero can't be in hundreds place. But let's assume that our number could start with zero.
  
  The formula to find sum of all numbers in a permutation is
  
  111 * no of ways numbers can be formed for a number at given position * sum of all given digits
  
  No of 1 s depends on number of digits
  
  So,the answer us
  
  111 * 20 *(0+1+2+3+4+5) = 33300
  
  We got 20 as follows..if we have 0 in units place we can form a number in 4*5 ways...this is for all numbers. So we have substituted 20 in formula..hope u understood
  
  Now, this is not the final answer because we have included 0 in hundreds place...so we have to remove the sum of all numbers that starts with 0.
  
  This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5.. because 0 at first place makes it a 2 digit number.
  
  So the sum for this is 11 * 4 * (1+2+3+4+5).
  =660
  
  Hope u understood why we use 4..each number can be formed in 4*1 ways
  
  So, the final answer is 33300-660 = 32640
  
  
• 8 years agoHelpfull: Yes(45) No(6)
• total no of 3 digit numbers possible are 5*5*4=100
  sum of all 3digit no= 111*100(0+1+2+3+4+5)=11100*15=166500
• 8 years agoHelpfull: Yes(11) No(3)
• number of digit given is n(including 0)
  number formed by r digit
  formula
  sum of all r digit is
  n-1 p r-1(1111.....r times)(sum of digits)-n-2 p r-2(1111.......r-1 times)(sum of digits)
  here,
  n=6(0,1,2,3,4,5)
  r=3
  sum is
  5 p 3(111)(0+1+2+3+4+5) - 4 p 1(11)-(0+1+2+3+4+5)
  20(111)*15 -4(11)*15
  300(111)-60(11)
  32640
• 7 years agoHelpfull: Yes(4) No(0)
• There are 5*5*4=100 way
  If you pick one at random, the expected value of the first digit is (1+2+3+4+5)/5=3
  and the expected value of each other digit must be 12/5
  because the sum of all five digits is 0+1+2+3+4+5=3+5*12/8
  0+1+2+3+4=52+4⋅158. By linearity of expectation, the sum of all 100 numbers is 100(3*100+12/5*11)=
  32640 is ans
  
• 8 years agoHelpfull: Yes(2) No(2)
• we know that number of way to organise a number in a way..
  for above question..
  suppose we fix the 5 at the end of number_ _ 5.then we have to find the number of way to find the 2 digit number including 0 either they appear in tens,hundread,unit place..for 2 number number of way
  0,1,2,3,4,5=number of way =5*4=20.
  for total sum including 0 each number 20 time in unit or hundread place so...
  total sum(0+1+2+3+4+5)*(20+200+2000).
  for unit =20 for tenth =20*10=200
  for hundread 200*10=2000.
  total=(0+1+2+3+4+5)*(20+200+2000_)=33300.
  and subtract number statt with zero..
  fix the 5 at end and does not consider 0from 0,1,2,3,4,5.
  _ _5 number of way4*3=12 way.
  (1+2+3+4+5)(12+120)=1980.
  sum=33300-1980=31320.
• 8 years agoHelpfull: Yes(1) No(4)
• Total numbers that can be formed= 5*5*4=100
  How many time 0 will appear at unit place?
  5*4=20 (Since 0 is at unit place)
  How many times 1,2,3,4,5 will comes at unit place?
  Lets take 1 at unit place. So, 0 can not be at leading position.
  4*4=16
  Similarly, 2,3,4,5 will appear 16 times at units position.
  So, Sum of Unit place digit:
  16*(1+2+3+4+5)=16*15=240
  Same number of times all numbers will appear at tens place and hundreds place. So,
  100*240+10*240+1*240=24000+2400+240=26640
• 8 years agoHelpfull: Yes(1) No(4)
• 3 digit number without repetition formed as
  102, 103, 104, 105, 203, 204, 205, 304, 305, 405
  now we have to find sum of all above no...........
  102-> 2!*111*3
  103->2!*111*4
  .
  .
  .
  .405->2!*111*9
  now sum of all number will be=2!*111*(3+4+5+6+5+6+7+7+8+9)
  = 2!*111*60
  =13410
  this will be the answer
• 8 years agoHelpfull: Yes(1) No(2)
• @jaya why multiply with 111?
• 8 years agoHelpfull: Yes(0) No(1)
• @Anjali if repeatation is not allowed then answer will be 8, if allowed answer will be 12.
  no is 345,375,435,465,645,675,735,765
• 8 years agoHelpfull: Yes(0) No(0)
• If you fix 5 as the last digit, you see that there are 4⋅4⋅ways to complete the number. Thus, 5 appears 16 times as the last digit. By the same logic, if we enumerate all possible numbers using these 6 digits, each number appears 16 times in each of the 3 positions. That is, the digit 5 contributes addition of total numbers In total, we have
  (0+1+2+3+4+5)(16+160+1600)= 26640
• 8 years agoHelpfull: Yes(0) No(0)
• @jaya ganthan how 4 has come in two digit no
• 7 years agoHelpfull: Yes(0) No(0)
• total no of 3 digit numbers possible are 5*5*4=100 therefore n=100
  First term can be 102 =a
  Last Term can be 543 =l
  So using Sum=n/2[a+l]=100/2[102+543]
  ans=32250
• 7 years agoHelpfull: Yes(0) No(1)
• best solution @rajeev kumar.. simple and effective approach .
• 7 years agoHelpfull: Yes(0) No(0)
• sum of unit digits of all the no=(total no of digits that can be placed on hundred place)*(total no of digits that can be placed on tenth place)*(0+1+2+3+4+5)=4*4*15
  
  sum of tenth digits of all the no=(total no of digits that can be placed on hundred place)*(0+1+2+3+4+5)*(total no of digits that can be placed on unit place)=4*15*4
  
  sum of hundred digits of all the no=(1+2+3+4+5)(total no of digits that can be placed on hundred place)**(total no of digits that can be placed on unit place)=15*5*4
  
  note-[ 0 cant be placed on hundred place, hence0 not added in the last ] also [for this it is now 5*4 not 4*4]
  now you can add up and take carries .
  ans is 32640
• 6 years agoHelpfull: Yes(0) No(0)