a five digit number is formed using the digits 1 3 5 7 9 what is the sum of all possible values of this number
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We have 5 digit 1, 3, 5, 7, 9.
So, total number of 5 digit number will be 5!
Lets take one such number, 53179.
This number can be written as 5∗104+3∗103+1∗102+7∗10+9.
Now, think about particular position. A digit will come at particular position 4! times, because rest of the digit can be arranged in 4! times.
For example 5 will come at 103 position 4! times. It means when we will be summing all these digit we will be having 5∗103, 4! times. it means 5∗103∗4! will be part of sum.
if we apply this analogy to all the digit and sum them, we will get this expression.
(104+103+102+10+1)∗(5+3+1+7+9)∗4!
=11111∗25∗4!
=11111∗600
=6666600
So, total number of 5 digit number will be 5!
Lets take one such number, 53179.
This number can be written as 5∗104+3∗103+1∗102+7∗10+9.
Now, think about particular position. A digit will come at particular position 4! times, because rest of the digit can be arranged in 4! times.
For example 5 will come at 103 position 4! times. It means when we will be summing all these digit we will be having 5∗103, 4! times. it means 5∗103∗4! will be part of sum.
if we apply this analogy to all the digit and sum them, we will get this expression.
(104+103+102+10+1)∗(5+3+1+7+9)∗4!
=11111∗25∗4!
=11111∗600
=6666600
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