find the sum of all number greater than 10000 that can be formed using digits 0,2,3,4,5 if repetition is not allowed.

• toal no. formed which is greater than 100000 =4*4*3*2*1
  =96;
  no. started with only 4 no. i.e 2,3,4,5
  sum of no=6(2+3+4+5)(4*10^4+3*10^3+3*10^2+3*10+3);
  =6*14*43333
  =3639972
• 9 years agoHelpfull: Yes(6) No(1)
• 0,2,3,4,5
  numbers must be greater than 10,000 and repitition not allowed
  so, we consider only 5 digit numbers
  _ _ _ _ _
  first digit do not start with 0
  so, for first digit we have 4! ways
  2nd digit also have 4! ways
  (2 numbers are over)
  3rd digit-3! ways
  4th digit-2! ways
  5th-1!
  finally 4!*4!*3!*2!*1! ways
  =24*24*6*2*1
  =6912
• 9 years agoHelpfull: Yes(5) No(3)
• first num does not starts with 0 so
  4*5*4*3*2*1=480(since repetation is not allowed)
• 9 years agoHelpfull: Yes(0) No(4)
• to find out sum of n-digit numbers formed by digit using x,y,z...upto n digits
  =(n-1)!*(x+y+z)*(111...n times )
  so let fix 2 at ten thousand place then we have 0,3,4,5 to make 4 digit numbers
  so total sum of numbers starting with 2 will be
  2*24*10000+3!(0+3+4+5)*1111=48000+6*12*1111=127992
  similary for 5
  5*24*10000+3!(0+3+4+2)*1111=120000+6*9*1111=179994
  similarly for 3
  3*24*10000+3!(0+2+4+5)*1111=72000+6*11*1111=145321
  similarly for 4
  4*24*10000+3!(0+3+2+5)*1111=96000+6*10*1111=162660
  so total sum=127992+179994+145321+162660=615697
• 9 years agoHelpfull: Yes(0) No(0)