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

6 years ago


Answers : (3)


Hi Rajat,

The answer is :

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

calculate to get the answer.

Please feel free to ask as many questions you have.


6 years ago

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

6 years ago


(Sum of all n digits)[n-1Pr -1 X (111...rtimes)]

i.e (1+2+3+4+5)(4p3x(1111)=15x4!x1111=399960

4 years ago

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