In question 1b, it says that the expected compression ratio should be 6/7. The original size would be 49 (7 bits for each of the 7 letters) - so that means that after compressing our size should be 42. BUT! how can it be 42, given that we have already 21 for sure? If we have 21 that means we need other 21 and that is not possible in any combination of 18 (the size of the [k,m] unit) or 7 (the bits size of a letter).
One of the answer is abcabca, which will be represented of course as a,b,c,[3,4]. That yields 39 bits.