Question

how many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even?difficulty level : easy

Answer 1

123456 is the answer for this question

Answer 2

given 6th digit even number , so last digit 2 or 4 or 6-> 3 ways
" 5th digit should be even...so there will be 2 ways(rep. not allowed)
so,therefore we get 5*4*3*2*2*3=720 ways
there are 6 digits.
say a, b, c, d, e, f

f=possible digits are 3 -(2,4,6)
e=possible 2 digits - (2,4,6)
and then rest...
d=possible 5 digits
c=possible 4 digits
b=possible 3 digits
a=possible 2 digits
the digits goes on decreasing since digits are not supposed 2 b repeated...
hence ans=>3*2*5*4*3*2=720...:)