The question asks for number of pairs (x,y) in which the x and y are 7 bit strings of 0 and 1. and they must differ in exactly one position
We can solve this using combinatorics approach. each position in the pair of strings has 4 choices
{00,01,10,11}
Now first select a position to keep the bit same in 7 ways since 7 bit string is given.
To keep this bit different we have two choices {01,10}.
Now we are left with 6 positions and they must be same, so now again for each of 6 bits we have two choices {00,11}
so 2^6 choices for 6 bits left.
So total no. of ways is 7*2*2^6 = 896