What is the sum of 5digit nos formed by digits 1,3,5,7,9?
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Answered by
1
Hey,
We have 5 digit 1, 3, 5, 7, 9.
So, total number of 5 digit number will be 5!5!
Lets take one such number, 53179.
This number can be written as 5∗104+3∗103+1∗102+7∗10+95∗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!4! times.
For example 5 will come at 103103 position 4!4!times. It means when we will be summing all these digit we will be having 5∗1035∗103, 4!4! times. it means 5∗103∗4!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!(104+103+102+10+1)∗(5+3+1+7+9)∗4!
=11111∗25∗4!=11111∗25∗4!
=11111∗600=11111∗600
=6666600
HOPE IT HELPS YOU:-))
We have 5 digit 1, 3, 5, 7, 9.
So, total number of 5 digit number will be 5!5!
Lets take one such number, 53179.
This number can be written as 5∗104+3∗103+1∗102+7∗10+95∗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!4! times.
For example 5 will come at 103103 position 4!4!times. It means when we will be summing all these digit we will be having 5∗1035∗103, 4!4! times. it means 5∗103∗4!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!(104+103+102+10+1)∗(5+3+1+7+9)∗4!
=11111∗25∗4!=11111∗25∗4!
=11111∗600=11111∗600
=6666600
HOPE IT HELPS YOU:-))
Answered by
0
HEY YOUR ANSWER :::;25
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