How many six digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the numbers should not repeat and the second last digit is even?

• here we have 7 numbers.we have 3 even numbers so that last digit can be arranged in 3 ways.second last digit can be arranged in 2 ways.the other digits can be arranged in 5,4,3,2 ways so the final answer is 3*2*5*4*3*2=720 ways
• 11 years agoHelpfull: Yes(47) No(4)
• 6*5*4*3*3*2
  second last disgit has 3 possibilities,so remaining 6 no.s can be placed in other vacancies.
• 11 years agoHelpfull: Yes(13) No(1)
• last digit can be filled in 3 ways...2,4,6
  second last digit cn be filled in 2ways...
  remaining 4 digits can be filled in 5p4
  total nubers= 5p4*2*3
• 11 years agoHelpfull: Yes(6) No(0)
• 6!=720 ways
• 11 years agoHelpfull: Yes(3) No(7)
• it will be 720 as it is solved in 5*4*3*2*2*3= 720
• 10 years agoHelpfull: Yes(3) No(0)
• suppose we have 6 blocks say.. _ _ _ _ _ _
  now according to the question last block will be even and second last block will also be even...so for 2nd block element out of three that is...2,4,6 one can be selected in 3c1 ways and for last block in 2c1 ways...and rest can be arranged in 5! ways
  
  and by multiplication rule...5!*3c1*2c1=720 ways is the required answer
• 11 years agoHelpfull: Yes(2) No(1)
• six digit even no means unit place must be even i.e.at unit place must be 1,3,5,7
  
  6c5*4!=6!/(5!*1!)*4!
  6*4!
  6*4*3*2*1=144
  
  answer=144
  
• 11 years agoHelpfull: Yes(1) No(10)
• 720.
  (5*4*3*2)*6
• 11 years agoHelpfull: Yes(1) No(0)
• no. of odd digits = 4 = (1,3,5,7)
  no. of even digits = 3 = (2,4,6)
  no. should be 6 digit no
  so,
  (1st) (2nd) (3rd) (4th) (5th) (6th)
  1357(246)
  ways to arrange 3 even digits at the last 2 places = 3 P 2 = 3! = 6
  ways to arrange 4 odd digits at 1st 4 places = 4 P 4 = 4! = 24
  so,
  no. of ways of arranging such 6 digits = 6*24 = 144
• 10 years agoHelpfull: Yes(1) No(5)
• 2*3*4*5*2*3
• 10 years agoHelpfull: Yes(1) No(0)
• 6!=720 numbers can be formed
• 11 years agoHelpfull: Yes(0) No(7)
• here we have 3 even numbers so they are arranged in 3! ways, similarly the remaining 4 numbers are arranged in 4! ways so the answer is 3!*4!=144
• 11 years agoHelpfull: Yes(0) No(7)
• sinc the 2nd last digit is even, we have to arrange 2,4,6 in that place as 3p1 ways and the remaining 5 places are arranged with the remaing digits(6 digits) as 5p6, hence the required ans is 3p1+5p6= 3 ways...
• 11 years agoHelpfull: Yes(0) No(6)
• it should be 144 because we can fill the last digit with only 3 cases and last but one by 2 and remaining 4 places 4 digits ie 4! so finally solution is
  4!*2*3=144
• 11 years agoHelpfull: Yes(0) No(4)
• selecting the appropriate digits is
  3*4*3*2*1*2=144
• 10 years agoHelpfull: Yes(0) No(0)