Vriti Edustore
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Sorr there was a minute mistake in the above post.
The answer is
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all units places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all tens places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all hundreds places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all thousands places
Now,
Addition of all units places = 40500
Addition of all tens places = 40500
Addition of all hundreds places = 40500
Addition of all thousands places = 45000
So the ans wer is
49495500
If everything is correct than I am right but check it because I solved it quite quickly with less precautions.
(Sorry if it wrong and please inform me, I willcorrect it.)
the total no. formed with 0,1,2,3 will be
4! =24 ,to find the sum of these 24 no., we will find the sum of digits at unit's,ten's,hundred's and thousand's places in all these no.
consider the digits in the unit's places in all these no. Each of the digit 0,1,2,3 occurs in 3! (=6) times in the unit's place.
So,total for the digits in the unit's place in all the no.
=(0+1+2+3) X 3! =36
Since each of the digit 0,1,2,3 occurs in 3! (=6) times in any one of the remaining places.
So the sum of digits in the digits in the ten's,hundred's and thousand's places in all the no.
=(0+1+2+3) X 3! =36
Hence,the sum of all the no.=36(100+101+102+103)=39996
[I wish my ans. will be liked by you]
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The answer is
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all units places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all tens places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all hundreds places
(1+2+3+4+5+6+7+8+9)*10*10*9 = Addition of all thousands places
Now,
Addition of all units places = 40500
Addition of all tens places = 40500
Addition of all hundreds places = 40500
Addition of all thousands places = 45000
So the ans wer is
494955
If everything is correct than I am right but check it because I solved it quite quickly with less precautions.
(Sorry if it wrong and please inform me, I willcorrect it.)