How many DNA chains of length 3 have no C’s at all or have no T’s in the first position?

Please note that my solution differs from the authors’ one given at the end

There are \(3^3\) chains of length 3 with no C’s at all and \(3 \cdot 4^2\) chains of length 3 with no T in the first position. However, these two sets have some overlap (they’re not disjoint, i.e. in both sets we have DNA chains with no C’s and not initial T. eg. \(AAG\) or \(GAT\)) and we have counted those chains twice so we need to subtract them. The number of DNA chains in the overlap, with no C’s and no initial T, is \(2 \cdot 3^2\) (two choices for the initial position (no C or T), and 3 choices for the two subsequent positions (no C)), so finally we arrive at a total of:

\[3^3 + 3 \cdot 4^2 - 2 \cdot 3^2= 57\]

Authors’ answer: \(3^3 + 3 \cdot 4^3\)¹