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

Options

A. `(2+3+4+5)xx3!`
B. `(2+3+4+5)((10^(4)-1)/(10-1))`
C. `(2+3+4+5)xx3!xx((10^(4)-1)/(10-1))`
D. `xx3!xx((10^(4)-1)/(10-1))`

Explanation

The total number of numbers formed with the digits 2, 3, 4, 5 taken all at a time `=""^(4)P_(4)=4!""=24`.
To find the sum of these 24 numbers, we will find the sum of 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. Each of the digits 2, 3, 4, 5 occurs in `3!(=6)` times in the unit's place.
So, total for the digits in the unit's place in all the numbers
`=(2+3+4+5)xx3!`
Since each of the digits 2, 3, 4, 5 occures 3! times in any one of the remaining places. So, the sum of the digits in the ten's hundred's and thousand's places in all the numbers
`(2+3+4+5)xx3!`
Hence, the sum of all the numbers
`=(2+3+4+5)xx3!(10^(0)+10^(1)+10^(2)+10^(3))`
`=(2+3+4+5)xx3!xx((10^(4)-1)/(10-1))`