A 5-digit number is formed by the digits 2,4,5,6,8 (each digit used exactly once) . What is the probability that the number formed is divisible by 4 ?

• a no. is divisible by 4 if last two digit is divisible by 4
  so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
  last two places are fixed by 24,28,48,52,56,64,68,84
  so remaining 3 places can be filled in 3! ways for each
  total=8*3!=48=n(E)
  n(S)=5C4=120
  p=48/120=2/5
  
• 11 years agoHelpfull: Yes(46) No(3)
• numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
  remaining 3 places can be filled in 3! ways....total=8*9=48
  probability=(48/120)=2/5
• 11 years agoHelpfull: Yes(17) No(3)
• Arranging 4 or 8 at the end of the required digit to make it divisible by 4....2C1
  
  Remaining 4 places are to be filled by four numbers....4P4(as order is important)
  req answer=2C1*4P4/5P5 (5P5 is sample space)
  =2*4!/5!
  =48/120
  =2/5
  
  
• 10 years agoHelpfull: Yes(5) No(0)
• here all the below process is correct.but 5c4=5 but not 120
• 10 years agoHelpfull: Yes(4) No(0)
• actually all are written 5C4........................Its not 5C4 ...it is 5P4......
  
  in this problem we are arranging the digits not selecting.
  
  and 5C4 value is .........5 not 120
• 10 years agoHelpfull: Yes(4) No(0)
• a no. is divisible by 4 if last two digit is divisible by 4
  so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
  last two places are fixed by 24,28,48,52,56,64,68,84
  so remaining 3 places can be filled in 3! ways for each
  total=8*3!=48=n(E)
  n(S) = 5!
  so
  asn= n(E)/n(S)
  =48/120=2/5
  
• 9 years agoHelpfull: Yes(4) No(0)
• total=8*3!=48=n(E)
  n(S)=5C4=120
  p=48/120=2/5
• 10 years agoHelpfull: Yes(2) No(0)
• LAST 2 nos must be divisible by 4.
  24,28,48,52,56,64,68,84
  and rest 3 places can be filled = 3*2*1 ways(as no repetition)
  so total = 8*6=48
  5C4=120
  p=48/120.
  
• 10 years agoHelpfull: Yes(1) No(0)
• for the no. should be divisible by 4 the cases are 24,28,48,64,68 and 84 so no. of cases possible be 6X3! and total no. of cases be 5! so p=3/10
• 10 years agoHelpfull: Yes(1) No(1)
• 120 ways..
• 10 years agoHelpfull: Yes(1) No(1)
• 1/4
  Numbers ending with 24,28,48,84,56 will be divisible by 4.
  Total numbers ending with above last 2 digits combination=(3!)*5=30
  Total 5 digit Numbers formed with 5 digits without repetition=5!=120
  So probability=30/120=1/4
• 11 years agoHelpfull: Yes(0) No(11)
• 25648
  divisible by 4
• 10 years agoHelpfull: Yes(0) No(1)
• ans is : 48
• 9 years agoHelpfull: Yes(0) No(1)
• correct answer=48
  the condition of divisible of 4=the two unit digit divisible by 4
  so total no=24/28/48,52,56,64,68,84 so total=8
  accourding to question repeatation is not allow so
  3p3*8=48
• 8 years agoHelpfull: Yes(0) No(0)