Find the sum of all 4- digited numbers that can be formed using the digits 1,3,5,7,9.

Explanation

We know that the number of 4 digited numbers that can be formed using the digits 1,3,5,7,9 is `""^(5)P_(4)=120`. ltbr We have to find their sum. We first find the sum of the digits in the units place of all the 120 numbers. Put 1 in the units place.
`square square square 1`
The remaining 3 places can be filled with the remaining 4 digits in `""^(4)P_(3)` ways. Which means that there are `""^(4)P_(3)` number of 4 digited numbers with 1 in the units place. Similaly, each of the other digits 3,5,7,9 appears in the units place `""^(4)P_(3)` times. Hence, by adding all these digits of the units place, we get the sum of the digits in the units place.
`""^(4)P_(3)xx1+""^(4)P_(3)xx3+""^(4)P_(3)xx5+""^(4)P_(3)xx7+""^(4)P_(3)xx9`
`=""^(4)P_(3)(1+3+5+7+9)`
`=""^(4)P_(3)(25)`.
Similarly, we get the sum of all digits in 10's place also as `""^(4)P_(3)xx25`. Since it is in 10's place, its value is
`""^(4)P_(3)xx25xx10`.
Like this the values of the sum of the digits in 100's place and 1000's place are respetively,
`""^(4)P_(3)xx25xx1000 and ""^(4)P_(3)xx25xx1000`.
On adding all these sums, we get the sum of all the 4 digited numbers formed by using the digits 1,3,5,7,9. Hence the required sum is
`""^(4)P_(3)xx25xx1+""^(4)P_(3)xx25xx10+""^(4)P_(3)xx25xx100+""^(4)P_(3)xx25xx1000`
`=""^(4)P_(3)xx25xx1111`
`=24xx25xx1111=6,66,600.""^(3)P_(2).`