Find the sun of all 4 digit nosthat can be formed using the digits 2 3 5 8 No digit

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
11 years agoHelpfull: Yes(4) No(0)
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
11 years agoHelpfull: Yes(2) No(0)
Answer should be 119988
11 years agoHelpfull: Yes(1) No(0)
(2,3,5,8) in 4! ways..each number comes in 6 times..
so 6*(2+3+5+8)*(1+10+100+1000) =119988
11 years agoHelpfull: Yes(1) No(0)
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).
11 years agoHelpfull: Yes(0) No(0)
the answer is 4! (4 factorial) .. that is 4*3*2*1 = 24
(applicable for series not getting repeated)
11 years agoHelpfull: Yes(0) No(1)

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