find the sum of the all the number formed by 2,4,6 and 8 without reputation.Number may be of any digit like 2,24,684,4862.

• (2+4+8+6)*3!*1111 = 133320 (CORRECT ANSWER)..
• 11 years agoHelpfull: Yes(17) No(7)
• 2+4+6+8=20
  24+26+28+46+48+68=240
  42+62+82+64+84+86=420
  246+264+426+462+624+642=2664
  268+286+628+682+826+862=3552
  248+284+428+482+824+842=3108
  468+486+648+684+846+864=3996
  2468+2486+2648+2684+2846+2864=15996
  4268+4286+4628+4682+4826+4862=27552
  6248+6284+6428+6482+6824+6842=39108
  8246+8264+8426+8462+8624+8642=50664
  so,20+240+420+2664+3552+3108+3996+15996+27552+39108+50664=147320
• 11 years agoHelpfull: Yes(15) No(7)
• 1 digit no sum=2+4+6+8=20
  2 digit no sum=22*3+44*3+66*3+88*3=3*220
  3 digit no sum=222*6+444*6+666*6+888*6=6*2220
  4 digit no sum=2222*6+4444*6+6666*6+8888*6=6*22220
  now add all(22220+2220)*6+660+20=147320
  explanation:
  for 1 digit it is simple
  for 2 digit no we know if we place 2 at 10th position then the other 1 digit may be taken from rest(4,6,8)and if we place 2 at 1th position then also the 10th position filled by one of from(4,6,8). thus 20*3+2*3=22*3 and same thing will happen for 4 6 8 thus for 2 digit sum will be 22*3+44*3+66*3+88*3=3*220
  ....for 3 digit we can choose 3 out of 4 by 4 ways,no made by one of these groups(246,468,268,248)we can see every digit comes in three groups.we take 2 first for 2 at 100th this thing will happen 2 time in one group that is total 6 time bcoz there are 3 groups simillarly for 10th place and for 1th place thus
  200*6+20*6+2*6=222*6 thus total 222*6+444*6+666*6+888*6=6*2220
  .....lastly for 4 digit 2468 if we place 2 at any position other can arranged by 3!=6 different way that means 2000*6+200*6+20*6+2*6=2222*6 same for all thus
  2222*6+4444*6+6666*6+8888*6=6*22220
  
• 11 years agoHelpfull: Yes(15) No(10)
• Let,
  No of digits : m
  No of digits for which we are finding he sum : n
  sum of digits : sod
  
  Suppose we are finding all 3 digit no from 4 digits(2,4,6,8):
  
  First we will find for unit place : how many times 2 will come :
  2 is fix now, so other arrangement = 3P2 i.e 6 (m-1Pn-1)
  similarlly all digits will come at unit place.
  similarlly for tens place ...
  
  So Sum of n digits from m digits,S :
  = (1*sod* m-1P n-1)+
  (10*sod* m-1P n-1) +
  (100 * sod * m-1P n-1) +
  ...
  ...
  (10^n-1 * sod * m-1P n-1)
  S(m,n) = sod* m-1 P n-1 * 111...1(n times)
  
  _________________________________________________
  puting this in this question :
  
  Sum of 4 digit numbers = (2+4+6+8)* 3P3 * (1111) = 20*6*1111 = 133320
  Sum of 3 digit numbers = (2+4+6+8)* 3P2 * (111) = 20*6*111 = 13320
  Sum of 2 digit numbers = (2+4+6+8)* 3P1 * (11) = 20*3*11 = 660
  Sum of 1 digit numbers = (2+4+6+8)* 3P0 * (1) = 20*1*1 = 20
  
  Adding All ,
  Sum = 147320
  
• 11 years agoHelpfull: Yes(13) No(3)
• there is a formula to solve such type of problem :
  (sum of all digits) * 1111...(r times) * (n-1)P(r-1)
  
  where
  P - Permutation
  n - Total given digits
  r - No. of digits in a number
  
  So solution like as following:
  
  1. for 2+4+6+8 => n=4,r=1 -> (2+4+6+8)*1*(4-1)P(1-1) = 20
  
  2. for 24+26+28+46+48+68+42+62+82+64+84+86 => n=4,r=2 -> (2+4+6+8)*11*(4-1)P(2-1) = 660
  
  3. similarly n=4,r=3 -> (2+4+6+8)*111*(4-1)P(3-1) = 13320
  
  4. and n=4,r=4 -> (2+4+6+8)*1111*(4-1)P(4-1) = 133320
  
  So to
• 10 years agoHelpfull: Yes(10) No(1)
• 143720 will be sum of all the numbers formed by 2,4,6 and 8 without reputation.
• 11 years agoHelpfull: Yes(2) No(0)
• is there any other method to solve this q???
• 11 years agoHelpfull: Yes(1) No(1)
• 147320 is correct
  
• 11 years agoHelpfull: Yes(1) No(0)