4. How many 6 digit even numbers can be formed by using 1,2,3,4,5,6 and 7 such that the second last digit is even?

• if repetation is allowed then 7*7*7*7*3*3=21609
  if repetation is not allowed then 2*3*4*5*2*3=720
• 10 years agoHelpfull: Yes(45) No(1)
• 5*4*3*2*2*3=720
  last digit can be 2,4,6= 3 ways
  2nd last digit = 2 ways (among 2,4,6 one is already selected so 2 ways)
  now we have 4 places and 5 digits left so 5*4*3*2 ways
  hence 5*4*3*2*2*3= 720 ways
• 10 years agoHelpfull: Yes(44) No(10)
• 7*7*7*7*3*3 = 21609
• 10 years agoHelpfull: Yes(36) No(13)
• total there are 7 digits,with 3 even numbers
  for a number to be even last digit should be even
  hence last digit can be filled in 3ways
  2nd last sholud be even that can be filled in 2 ways
  remaining 4places can be filled by remaining 5 digits=5! ways
  total ways=5!*2*3=720ways
• 10 years agoHelpfull: Yes(8) No(3)
• if repeatition is allowed 7*7*7*7*3*3
  if repeatition is not allowed 5*4*3*2*2*3
• 10 years agoHelpfull: Yes(4) No(0)
• 2*3*4*5*3*6=2160
  5th position can be filled in 3 ways only
  
• 10 years agoHelpfull: Yes(2) No(0)
• 7*7*7*7*3*7
  
• 10 years agoHelpfull: Yes(2) No(4)
• _ _ _ _ _ _ there are 6 position. To get even no unit digit position should contain even no i.e. 2 or 4 or 6. Let at unit digit position 2 is there..so tharere are other 6 nos which will be placed at 5 position as it's nt mentioned that no repeatation is allowed so for 2 the ans is 5!..same ans for 4 & 6...
  The final ans will be 3*5!..
• 10 years agoHelpfull: Yes(1) No(8)
• 7*7*7*7*3*3 = 21609
• 10 years agoHelpfull: Yes(0) No(1)
• sorry,7*7*7*7*3*3
  
• 10 years agoHelpfull: Yes(0) No(0)