find the no. of ways in which a 6 digit no. can be formed from 1,2,3,4,5,6,7 such that the second last digit is always even.

• second last digit has 3 options.
  then first digit will have 6 options..... 2nd digit has 5 options... so on
  
  so total possible numbers = 3*6*5*4*3*2= 2160
• 12 years agoHelpfull: Yes(27) No(3)
• if no. can be repeated then,
  ans is (7^5)*3=50421
  and if no. can't be repeated then,
  ans is 6*5*4*3*3*2=2160
• 12 years agoHelpfull: Yes(9) No(1)
• 6!*3=2160
  fix 2,4,6 at 2nd last then permute remaing 6
• 12 years agoHelpfull: Yes(2) No(1)
• 6!*3=2160if no cnt be repeated
• 12 years agoHelpfull: Yes(1) No(0)
• 50421 if repetition is allowed as 7*7*7*7*3*7
• 12 years agoHelpfull: Yes(1) No(0)
• i think 2160 is the answer for this
• 12 years agoHelpfull: Yes(0) No(0)