Q:

How many DNA sequences of length $3$ that have no C's at all or have no T's in the first position? 

How many DNA sequences of length $3$ that have no C's at all or have no T's in the first position? 
Here are my workings below but I am not sure if they are correct or not,
So there are $4$ letters of the DNA sequence, CTGA
So considering DNA sequences of length $3$ that have no C's at all,
the first position can be filled by $3$ options, the second by $3$ options and the last by $3$ options, so therefore $3\times3\times3=27$
Considering DNA sequences of length $3$ that have no T's in the first position,
the first position can be filled by $3$ options, the second by $4$ options and the last by $4$ options, so therefore $3\times4\times4= 48$
So combining both cases, there are $48+27$ DNA sequences.  

A:

This can be done using the inclusion-exclusion principle.  If you want to find the number of things that are X or Y, you add the number of Xs to the number of Ys, then subtract the number of things that are both X and Y, since they were counted twice. 

There are $3\times3\times3 = 27$ sequences have no Cs. 
There are $3\times4\times4 = 48$ sequences that don't begin with T.
There are $2\times3\times3 = 18$ sequences that don't begin with T and have no Cs. 

So the total number is $27 + 48 - 18 = 57$.
As an alternate method, you could also count the number that have a T in the first position and at least one C (i.e., the ones that aren't in your solution set).  There are 7 of these.  Since there are 64 possible sequences, that means the number that are in your solution set is $64 - 7 = 57$, which agrees with the above.