A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
1.6666600
2.6666660
3.6666666
4.None

• Total Combinations without repetation is 5!=120.
  smallest no = 13579 largest no = 97531.
  take avg. = ((13579+97531)/2)*120= 6666600
  ie ans is 1.
• 10 years agoHelpfull: Yes(108) No(1)
• Ans: 6666600
  
  Explanation:
  Given digits are 1,3,5,7,9
  
  I am writing the numbers are in ascending form.. from given digits, numbers should be:
  
  97531 + 97513 + 97351 + ........+ 13759 + 13597 + 13579
  
  Now we 'll add first and last term of above sequence and again we 'll add second and second last term and so on...
  we 'll get
  97531 + 13579 = 111110
  97513 + 13597 = 111110
  97351 + 13759 = 111110
  ..
  ..
  Upto middle position of the above sequence.
  Now we see every time we are getting 111110
  
  now i have to find the how many possible number :
  ==> 5*4*3*2*1 = 120
  there are 120 numbers
  then the middle position is: 120/2 = 60
  
  so, sum of all possible numbers are....
  60*111110 = 6666600 Ans.
  
• 10 years agoHelpfull: Yes(27) No(2)
• 4!(1+3+5+7+9+)*11111
• 10 years agoHelpfull: Yes(14) No(4)
• Sum of arrangement of such numbers are given by:(n-1)!*n times 1*(sum of digits)
  here n = 5 and (n-1)! = 4! = 24..
  sum of (first 5 odd) digits = n^2 = 25
  thus 24*11111*25 = 6666600..option 1)
• 10 years agoHelpfull: Yes(9) No(0)
• ans is 1.6666600
  
  total possible no without replacement 5!=120
  
  how many time 1,3,5,7,9 occur at unit place
  24 times
  _ _ _ _ 1 (1*24 => 24)
  _ _ _ _ 3 (3*24 => 72)
  _ _ _ _ 5 (5*24 => 120)
  _ _ _ _ 7 (7*24 => 168)
  _ _ _ _ 9 (9*24 => 216)
  
  24+72+120+168+216 = 600 {Extract '0' from 600 for sum of all possible no acting as a unit place..
  so remain digit is 60.....Carry-1}
  
  again how many time 1,3,5,7,9 occur at ten place
  6 times
  _ _ _ 1 (1*6 => 6)
  _ _ _ 3 (3*6 => 18)
  _ _ _ 5 (5*6 => 30)
  _ _ _ 7 (7*6 => 42)
  _ _ _ 9 (9*6 => 54)
  
  6+18+30+42+54 => 150
  now add 150 to carry 60
  150+60 = 210...{Extract '0' from 210 for sum of all possible no acting as a Ten place..so remain digit is 21.....Carry-2}
  
  
  doing in this way i got my ans 6666600.....
  
  
  
  
• 10 years agoHelpfull: Yes(4) No(3)
• sum of all such possible numbers formula=n*(n-1)!*11111
  here n=5=5*4!*11111=6666600
• 10 years agoHelpfull: Yes(2) No(3)
• If you assume that any digit is in fixed position, then the remaining four digits can be arranged in 4! = 24 ways.
  
  So, each of the 5 digit will appear in each of the five places 24 times. So, the sum of the digits in each position is 24(1+3+5+7+9) = 600
  
  The sum of all such numbers will be 600(1 + 10 + 100 + 1000 + 10000) = 6666600
• 6 years agoHelpfull: Yes(0) No(0)