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`        Find the sum of all 4 digit numbers (without repetition of digits) formed using the digits 1to5.`
7 years ago

22 Points
```										Hi Rajat,
There are ( 5*5*5*5 = 625 ) 4 digit numbers fromed using the digits 1 to 5.
Now for 1000 place , each digit can come 125 times ( thus giving it total of 125 * 5 = 625 Numbers )
Thus sum for 1000 place will be ( 1000 * ( 1 + 2 + 3 + 4 + 5 ) * 125 ) = 1000*15*125
Similarly for 100 place each digit can come 125 times.
Same happens for 10th place and unit place
Thus total sum will be ( 1000 + 100 + 10 + 1 ) * 15 * 125
Puneet
```
7 years ago
Rajat Goel
31 Points
```										sorry for disapproving your answer. But the answer given in my answer booklet is 399960 while the solution you gave results out to be 2083125..
But still thanks a lot for helping me as when i tried this question this time using your method i find out right answer. you have just committed a mistake that is you did the ques with '4 digits number with repetitiom' while it is "without repition is given".

Thanks a lot
```
7 years ago
Magisetty Harikrushna
18 Points
```										Answer:
(Sum of all n digits)[n-1Pr -1 X (111...rtimes)]
i.e (1+2+3+4+5)(4p3x(1111)=15x4!x1111=399960
```
5 years ago
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