What is the prob of 4 digit no formed with 1 to 5 digit to be divisible by its unit place
1 96
2 216

• 4*3*2=24(when the last digit is 1 and it is divisible by 1)
  4*3*2=24(when the last digit is 2 and it is divisible by 2)
  4*3*2=24(when the last digit is 5 and it is divisible by 5)
  3*2 =6(when last 2 digits are 24 to check it is divisible be 4)
  total=78(as no digit wil be divisible by 3 keeping 3 in units place)
  
  if digits are not repeated this should be the solution
• 10 years agoHelpfull: Yes(12) No(11)
• i think....
  5*4*3*2=120(for checking divisibility by 1 bcos 1 is divisible by all no.s)
  3*2*1*2=12(for checking divisibility by 2 ,took [2,4] as a last digit)
  4*3*2*1=24(for checking divisibility by 5)
  2*1*3=6(for checking divisibility by be 4,took 3 digits that is [2,3,4])
  4*3*2*1=24(for checking divisibility by 3(sum of all digits is divisible by 3) so we will only take 4 digits i.e[1,2,4,5])
  
  so total will be:-186
  
• 10 years agoHelpfull: Yes(0) No(2)