How many 6-digit even numbers can be formed from the digits 1,2,3,4,5,6 and 7 so that digits should not repeat and the second last digit is even?
a)720
b)320
c)2160
d)6480

• EDIT in the above answer**
  ------
  5P4 * 6 since there are 6 cases
  
  5P4*6=720
• 10 years agoHelpfull: Yes(24) No(9)
• ans is 6*5*4*3*3*2=2160
  the second last digit is 2,4,6 only, so there is 3 posibilites to fill up the 3rd place..we have total 7 num, so 6th place can be fill with 6number, 5th place fill with 5 number and so on..
• 10 years agoHelpfull: Yes(16) No(4)
• I hope this is the answer!
  To satisfy the above conditions as in the question..(the second last and last number shuld be even)
  6 cases
  _ _ _ _ 42
  _ _ _ _ 62
  _ _ _ _ 24
  _ _ _ _ 64
  _ _ _ _ 26
  _ _ _ _ 46
  
  So 2 digits are already set as above
  Now take the first example. _ _ _ _ 4 2
  here the remaining digits that can be filled are 1,3,5,6,7
  So 5P3
  
  Since 6 such cases are there 5P3*6=720
• 10 years agoHelpfull: Yes(11) No(13)
• As the second last digit is even and the number is even, So that we have to fill up the last two digits with even numbers i.e. 3P2 .
  And the remaining four positions we can fill up in 5*4*3*2 ways.
  So that our required answer is 3P2*5*4*3*2 = 6*5*4*3*2 = 720
• 10 years agoHelpfull: Yes(9) No(5)
• positions 1 2 3 4 5 6
  
  5th should be even so posibility is 2,4,6=3
  other 5 can be any among 6 6p5=6!
  ans=6!*3=2160
  
• 10 years agoHelpfull: Yes(5) No(2)
• ans is 720 expl: for unit digit we have three options 2,4,6 then for second last 2 options left as one out of 2,4,6 used in unit digit then for remaining 5 places 5*4*3*2*1 so the ans is 5*4*3*2*2*3=720
• 10 years agoHelpfull: Yes(4) No(4)
• 3c2*2!*5*4*3*2=720
• 10 years agoHelpfull: Yes(0) No(3)
• 5!*3*2=720
  
• 7 years agoHelpfull: Yes(0) No(1)
• So according to question.......... making 6 digit even number.
  ....
  ..
  .
  .
  .
  2nd last digit is even( i.e 5th position must even)
  
  (1th)_____ (2nd) ______ (3rd) ______ (4th) _______ (5th) Even (6th)___________-
  .
  .
  .we have 3 even number........... Now number of way selecting 3 even number for one place( 5th position) = 3C1=3
  .
  .
  Remaining place( 1st, 2nd, 3rd, 4th, 6th)
  Total number remaining is six because out of 7 number one must be fixed at 5th place( only even number)
  ..
  .
  .
  number of arranging 6 number in 5 vacant place is = 6P5 = 6*5*4*3*2*1
  .
  .
  Now total number of digit form is = 6P5*3=2160
  ....
  AS per position (1th)6 (2nd)5 (3rd)4 (4th)3 (5th)( 3c1=3) (6th)2 (7th)1
  == 6*5*4*3*3*2*1= 2160
• 7 years agoHelpfull: Yes(0) No(1)