A software engineer creates a LAN game where an 8 digit code made up of 1,2,3,4,5,6,
7,8 has to be decided on as a universal code. There is a condition that each number has to be
Used and no number can be repeated. What is the probability that first 4 digits of the code are even number?
a)1/70 b)1/840 c)1/8 d)1/40320

• total no of ways is 8!.
  for finding even nos at last 4 digit s ways are 4!
  and that for odd is 4!
  so it can be done in 4*3*2*1*4*3*2*1/8*7*6*5*4*3*2*1
  ie 4!*4!/8!
• 8 years agoHelpfull: Yes(20) No(0)
• so it can be done in 4*3*2*1*4*3*2*1/8*7*6*5*4*3*2*1
  ie 4!*4!/8!
• 8 years agoHelpfull: Yes(9) No(0)
• option:a is correct
• 8 years agoHelpfull: Yes(5) No(2)
• a) 1/70
  8!/4!*(8-4)!
• 7 years agoHelpfull: Yes(3) No(0)
• a)1/70
  Total no of possible ways are 8!
  The total number of possible ways to find even numbers at the last 4 digits are 4!
  The total number of possible ways to find odd numbers at the last 4 digits are 4!
  So, the probability that the first 4 digits of the code are even number = 4!*4!/8! = 1/70
• 6 years agoHelpfull: Yes(2) No(0)
• 4*3*2*1*4*3*2*1/8*7*6*5*4*3*2*1
  ie 4!*4!/8!
  so the answer will be (a)1/70
• 6 years agoHelpfull: Yes(1) No(0)
• 5,3,4,?,38
• 8 years agoHelpfull: Yes(0) No(2)
• (4! *4!)/8!= 1/70
• 5 years agoHelpfull: Yes(0) No(0)