form all different number using 2,4,6,8 digits( eg: 2,46,8624) without repetition
find the sum of these numbers.

• 4 digits (2,4,6,8) so sum is
  (4-1)factorial*(2+4+6+8)*1111 = 133320
• 9 years agoHelpfull: Yes(15) No(2)
• 2+4+6+8=20
  24+26+28+42+46+48+62+64+68+82+84+86=660
  246+248+264......+864=13320
  2468+2486+2648+2684+.....=133320
  total: 147320
• 9 years agoHelpfull: Yes(14) No(13)
• (2+4+6+8)(1111)*3!+[(2+4+6)+(4+6+8)+(2+6+8)](111)*2!+[(2+4)+(4+6)+(2+6)+(2+8)](11)+2+4+6+8
  
  ans is 133320+13320+660+20= 147320
• 9 years agoHelpfull: Yes(5) No(2)
• The required numbers are as follows;
  
  8642
  8624
  8462
  8426
  8246
  8264
  6842
  6824
  6482
  6428
  6248
  6284
  4682
  4628
  4862
  4826
  4286
  4268
  2648
  2684
  2468
  2486
  2846
  2864
  
  sum 133320
• 8 years agoHelpfull: Yes(2) No(1)
• (2+4+6+8)(1111)*3+[(2+4+6)+(4+6+8)+(2+6+8)](111)*2+[(2+4)+(4+6)+(2+6)+(2+8)](11)+2+4+6+8
  
  ==666660+10212+374+20=77266
• 9 years agoHelpfull: Yes(1) No(4)
• First condition for single digits = 4
  Second for two digit = 4*3 = 12
  Third for three = 4*3*2 = 24
  Fourth for four digit = 4*3*2*1 = 24
  
  total no. will be = 24+24+12+4 = 64
• 9 years agoHelpfull: Yes(0) No(6)
• what is the answer pls explain
• 9 years agoHelpfull: Yes(0) No(0)
• 4 digit number without repetition
  so it will be 4*3*2*1=24
• 9 years agoHelpfull: Yes(0) No(0)
• 13320 is an answer
• 9 years agoHelpfull: Yes(0) No(0)
• The answer is 133320+124432+660+20= 268432
• 9 years agoHelpfull: Yes(0) No(0)
• same solution as abhishek singh .
  for single digit sum is = 20
  so on.
• 9 years agoHelpfull: Yes(0) No(0)