Q. A five digit number is formed 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 ?

• Sum = 6666600
  
  Total of diff numbers possible = 5! --> 5*4*3*2*1 = 120
  
  Smallest no = 13579
  Largest no = 97531
  Average of 120 numbers =(13579+97531)/2
  
  Sum of 120 nos = 111110/2 * 120
  = 111110*60
  = 6666600
• 12 years agoHelpfull: Yes(20) No(8)
• 25*24( 10000+1000+100+10+1) = 600* 11111= 6666600
• 12 years agoHelpfull: Yes(7) No(5)
• The possible 5 diget numbers formed is
  5x4x3x2x1=120 numbers
  The number of times each diget exist
  120/5 = 24 x each diget in a column
  total is
  24x(9+7+5+3+1) = 600
  sum total as 6666600
• 12 years agoHelpfull: Yes(6) No(11)
• For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.
  
  Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.
  
  120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...
  
  This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.
  
  We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...
  
  So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:
  
  Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
  +
  Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
  +
  Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
  +
  Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
  +
  Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)
  
  = 6,666,600
• 8 years agoHelpfull: Yes(4) No(0)
• Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times.
  Hence the sum of digits in each position = 24(1 + 3 + 5 + 7 + 9) = 600.
  So, the sum of all numbers
  =600(1+10+100+1000+10000)=6666600
• 8 years agoHelpfull: Yes(3) No(0)
• Sum of nmbrs= 25.
  There are total 5 diffrnt nmbrs given..So
  (N-1)!=24
  25×24×11111=6666600
• 9 years agoHelpfull: Yes(0) No(2)