How many no can be formed using digits 1,2,3,4,5,6,7,8,9 such that they are in increasing order(eg. 12345 ,578)

• If you start with 9, you can only do the one digit. If you start with 8, you can do 8, or 89. If you start with 7, you can do 7, 78, 79, or 789. If you start with 6, you can do 6, 67, 68, 69, 678, 679, 689, or 6789. I'll do one more. If you start with 5, you can do 5, 56, 57, 58, 59, 567, 568, 569, 578, 579, 589, 5678, 5679, 5689, 5789, 56789.
  so when we start with 9,8,7,6,5...upto 1 then we have 1,2,4,8,16,32,....256 ways
  respectively so add all these ways we get 2^0+2^1+...2^8=511
• 11 years agoHelpfull: Yes(24) No(4)
• 2digit+3digit+.....+9dgit
  
  9*8/2+ 9*8*7/6+ 9*8*7*6/24+ 9*8*7*6*5/120+.....=502
• 11 years agoHelpfull: Yes(7) No(2)
• 9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9=502
• 11 years agoHelpfull: Yes(4) No(3)
• 9c0+9c1+9c2+9c3+9c4+9c5+9c6+pc7+9c8+9c9= 2^9-1 =511.
• 11 years agoHelpfull: Yes(4) No(1)
• 9*8/2 + 8*7*9/2 + 7*6*9/2 + 6*5*9/2 + 5*4*9/2 + 4*3*9/2 + 3*2*9/2 + 2*1*9/2 = 766
• 11 years agoHelpfull: Yes(2) No(8)
• 8*9/2+7*8*9/2+6*7*8*9/2+5*6*7*8*9/2+4*5*6*7*8*9/2+3*4*5*6*7*8*9/2+
  2digi+3digits+4digits +5 digits +6 digits +7 digits +
  +2*3*4*5*6*7*8*9/2+1*2*3*4*5*6*7*8*9/2
  +8 digits +9 digits
• 11 years agoHelpfull: Yes(1) No(3)
• answer 520
• 11 years agoHelpfull: Yes(0) No(9)
• Could you elaborate your answer.
• 11 years agoHelpfull: Yes(0) No(1)
• use simple formula for every digit as-2^9-d where d is digit
  2^9-1+2^9-2+2^9-3............2^9-9=256+128+64+32+16+8+4+2+1=511
• 8 years agoHelpfull: Yes(0) No(0)