Find the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time.

Explanation

The total number of numbers formed with the digits 2,3,4,5 taken all at a time is equal to the number of arrangements of 4 digits, taken all at a time, i.e., `.^(4)P_(4)=4!=24`.
To find the sum of these 24 numbers, we have to find the sum of the digits at unit's, ten's, hundred's, and thousand's places in all these numbers.
Consider the digits in the unit's places in all these numbers.
If 2 is the digit in unit's place, the remaining three places can be filled in 3! ways or we can say 2 occurs in unit's place 3! (=6) times. Similarly, each digit occurs 6 times.
So, the total sum of the digits in the unit's place in all these number is `(2+3+4+5)xx3!=84`.
Similarly, the sum of digits is 84 in ten's, hundred's, and thousand's places.
Hence, the sum of all the numbers is `84(10^(0)+10^(1)+10^(2)+10^(3))=93324`.