Question

How many different numbers (without repetition of digits) can be formed from the digits $1,3,5,7,9$ when taken all at a time and what is their sum?

Answer 1

$1,3,5,7,9.$ five digits.
No. of numbers $=5!=120.$
We have to find the sum of these $120$ numbers.
Suppose 9 is in the unit place then the remaining 4 can be arranged in $4!=24$ ways. Similarly other digits can occupy the first place in 24 ways.
Hence sum due to the unit place of all the 120 numbers
$=24(9+7+5+3+1)$ units.
$=24\times 25$ units = 600 units
Again suppose 9 is in the 2nd place i.e. ten's place and it will be so in 24 numbers. Similarly each digit will be in ten's place 24 times. Hence the sum of digits due to ten's place of all the 120 numbers is
$24(9+7+5+3+1)$ tens
$=24\times 25tens=600tens=6000$.
Proceeding exactly for hundreds, thousands and ten thousands, we have the sum of the numbers
$=600(1+10+100+1000+10000)$
$=600\left(11111\right)=6666600.$