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

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.

Puneet

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

Answer:

(Sum of all n digits)[^n-1P_{r -1} X (111...rtimes)]

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