Latest TCS Aptitude Question SOLUTION: Find the sun of all 4 digit nosthat can be formed using the digits 2,3,5,8 .No digit is to be repeated?

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Find the sun of all 4 digit nosthat can be formed using the digits 2,3,5,8 .No digit is to be repeated?

Asked In TCS (10 years ago)

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Shashi Kumar Gupta  (10 years ago)

answer is 119988..

formula is..
sum = (n-1)!*(sum of n given digit)*(1111...n times)
i.e here n=4 so,
sum = (4-1)!*(2+3+5+8)*1111 =119988

Is this solution Helpfull? Yes (24) | No   

amit kumar  (10 years ago)

Total no. of 4 digit=24
and every digit comes 6 times at every place
so 8*6+5*6+3*6+2*6=108
Hence sum of all digit=
108000
10800
1080
108
--------------
119988

Is this solution Helpfull? Yes (4) | No   

rohit kumar  (10 years ago)

sum of all the numbers which can be formed by n diguts without repitition is,
(n-1)!*(sum of all the digits)*(1111....n times)
so,
sum=(4-1)!*(2+3+5+8)*(1111)=3!*18*1111
=6*18*1111=119988

Is this solution Helpfull? Yes (2) | No   

lucky  (10 years ago)

Answer should be 119988

Is this solution Helpfull? Yes (1) | No   

NARAYANA SWAMY DAMMALA  (10 years ago)

(2,3,5,8) in 4! ways..each number comes in 6 times..
so 6*(2+3+5+8)*(1+10+100+1000) =119988

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Zahiruddin ansari  (10 years ago)

There r 4 nos,so no of ways that can be formed 4 digit no. is:-4*3*2*1=24=4!
but here is sum of all 4 digit nos..
logic is that
6 ways which have 1 at unit place.....> 3_ 2_ 1_ (1_)unit place
so 3*2*1=6ways
lly,6 ways which have 2 at unit place and so on...
so 6(2+3+5+8)=108 i.e 1*108 at unit place
Now at ten's place
it is:=10*108 and so on...
the sum is:-(1000*108+100*108+10*108+1*108)
=108(1000+100+10+1)
=108*1111
=119988
and by the formula,it will be 1111*6(2+3+5+8).

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Baarath Gopalkrishnan  (10 years ago)

the answer is 4! (4 factorial) .. that is 4*3*2*1 = 24
(applicable for series not getting repeated)

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