Option A and B says about number of possible gray codes of length n.
"A gray code is a ordering of all n length binary strings starting with 0^n and successive n length strings will differ by one bit and first and last string also will differ by one bit".
So let's see how many possible gray codes can be found when n=1,2
when n=1 there is only one possibility {0 , 1}
when n=2 there is two possibilities {00,01,11,10} , {00,10,11,01}. It is because the gray code must start with 0^n and we can only have 01 or 10 after 00, and that is why we got two possibilities.
So from above observation we can clearly reject Option A and B because number of possible gray code for a given 'n' can be odd(in case of n= 1 it is 1) and also can be even(in case of n= 2 it is 2).
So A and B can not be our option
Now lets check for Option C and D
Lets take a possible gray code for n = 3
{000, 100, 101, 111, 110, 010, 011, 001}
Here 000 and 110 are separated by 3 strings (100,101,111) but they differ by 2 bits So That makes option C and D false.
So from above observations it is clear that Option E is the right choice.