Q. Sum of all 120 numbers formed by using 1, 2, 3, 3, 3, and 4 is?
Options
1) 35555516
2) 35555520
3) 45555516
4) 34555606
5) 45555520
6) 45555506
7) 34562816
8) 48355520
9) 48355516
10) None of these

• total no. = 6!/3!=120
  in which 20 no.s starting with 1,2 & 4 each,total 60
  & there will be 3*20=60 no. starting with 3
  
  sum=111111*(20*1+20*2+60*3+20*4)=35555520
• 11 years agoHelpfull: Yes(40) No(2)
• total no. = 6!/3!=120
  now we have to check now many time a digit is repeated. as there are 6 digits we do 120/6=20
  that means every digit is repeated 20 times except digit 3 which is repeated 3*20=60 times.
  
  sum=(sum of the digits at the unit place)*(111111)
  = (20*1+20*2+60*3+20*4)*111111=35555520
• 11 years agoHelpfull: Yes(28) No(0)
• RAKESH,can u tll me y u hv taken 60*3 in the solution.
  
• 11 years agoHelpfull: Yes(3) No(2)
• 20*16(100000+10000+1000+100+10+1)=111111*320=35555520
  
• 11 years agoHelpfull: Yes(2) No(1)
• 111111*(20*1+20*2+60*3+20*4)
  =35555520
• 11 years agoHelpfull: Yes(1) No(4)
• 2) 35555520
• 11 years agoHelpfull: Yes(0) No(2)
• None of the above
  right answer is 111*(20*1+20*2+20*4+60*3)=35520
• 10 years agoHelpfull: Yes(0) No(3)