All the numbers possible using the digits ‘6’, ‘5’, ‘3’ and ‘2’ without repetition are written. Find the sum of all these numbers.
a) 76,659
b) 106,656
c) 156,659
d) Cannot be determined
Ans: b) 106,656
When one digit occupies a particular position, the three can be arranged in 3! = 6 ways. Thus, each digit will occur in each of the 4 places 69 times. So the sum of digits in each place is 6(6 + 5 + 3 + 2) = 96. So the required sum = 96(1 + 10 + 100 + 1000) = 106, 656.