A five-digit number is formed using the digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?(1993)
Keeping one digit in fixed position, another four can be arranged in 4 ! Ways = 24 ways. Thus each of 5 digit will occur in each of five places 4! Times. Hence, the sum of digits in each position is 24 (1+3+5+9) = 600. So, the sum of all numbers = 6000 (1+10+100+1000+10000) = 6666600.
Shortcut:
Let n = number of digits
Then, The sum of all possible numbers is given by (n)!(sum of all the digits)(1111….n times)= 5!(18)(11111)= 6666600