how many no.s more than 1000 and less than 10000 can be formed such that they are divisible by 5 and no 2 digits are repeated ?

• fist conside no. with unit digit 5
  so we can fill 3 remaning digits with 8*8*7=448
  now consider unoit digit as 0
  so we ca fill 3 remaning digits with 9*8*7=504
  so total no can be formed =952
  
• 11 years agoHelpfull: Yes(12) No(2)
• answer is 952
  
  here we take 2 cases
  case 1: 5 is in the units digit
  here msb can be filled in 8 ways (from 0 to 9 except 0 and 5)
  second digit can be filled in 8 ways and third in 7 ways
  so no. of ways = 8*8*7 = 448
  
  case 2: when 0 is in the units place
  here msb can be filled in 9 ways (all digits from 0 to 9 except 0)
  and second digit can be filled in 8 ways and third digit in 7 ways
  so total number of ways = 9*8*7 = 504
  
  so total no. of arrangements = 448+504 = 952
  
• 11 years agoHelpfull: Yes(4) No(0)
• _ _ _ _ _=_ _ _ 5 if it has to be divisible by 5. also 1st digit cant be 0. thus solution is 8*7*6=336
• 11 years agoHelpfull: Yes(0) No(3)