Can someone explain why the answer for 12 is d) and why the answer for 13 is b)? I'm trying to study for a test tomorrow but I'm looking over the answers for this practice test and I genuinely don't understand why these are the correct answers/how to even attempt these types of questions to begin with.
I would really appreciate some guidance.
Sorry I didn't see this in time to help you study for your exam, but maybe the answer will be of use to you in the future.
For number $12$, the idea is to see how we can build up an awesome string of length $n$ from shorter awesome strings. Note that every substring of an awesome string is awesome, so we can form an awesome string of length $n$ by adding a single character at the beginning of an awesome string of length $n-1$, and every awesome string of length $n$ arises in this way. If $\omega$ is awesome, so are $b\omega$ and $c\omega$, so that gives us $2A_{n-1}$ awesome strings of length $n$ that start with $b$ or $c$.
How many start with $a$? If the second character is $c$ then the last $n-2$ characters can be any awesome string, so that accounts for another $A_{n-2}$. The second character cannot be $a$, so we have to count the awesome strings starting with $ab$. Now the third character must be $c$, since $a$ and $b$ are prohibited, and then then remaining $n-3$ characters can be any awesome string.
Adding it all up, we get answer $(d)$.
You might ask, why don't we add characters at the end, instead of the beginning? I don't have an answer to that. In fact, I started by trying to do it that way, but I got a mess, and decided to try starting from the other end before trying to untangle things. Starting at the front turned out to be easy.
As to problem $13$, it's easy to compute $M(n,k)$, so I would do that, and see which answer worked. There are $\binom{n}{k}$ ways to choose the columns in which a $1$ will appear, and for each there are two ways to pick the row, so $$M(n,k)=\binom{n}{k}2^k$$