Combinations (counting sequence) of words I've been working on these two practice problems, I'm not really sure whether my solutions are right or not. 


*

*What is the counting sequence for all words on the $\{a, b, c\}$ that contain exactly one $a$? 
For this question I've come to the conclusion that the sequence is $\binom{n}{1}\cdot2^{n-1}$

*We have $\{a, b\}$, no consecutive "b"s are allowed.
How does the counting sequence begin (at least up to size $6$)? 
I think I should take the combination with no repetition approach, but I have no idea where to begin with.
A step by step guide for this problem is appreciated.
 A: For part 2: let $c_n$ denote the number of $n$ letter words satisfying your restraints. Let $w=l_1l_2\dots l_n$ be such a word, where $l_i\in\{a,b\}$ for $1\leq i\leq n$. If $l_1=a$, then this induces no restrictions on the remaining letters $l_2\dots l_n$. This means that $l_2\dots l_n$ is a word of length $n-1$ satisfying the repetition requirements, i.e. there are exactly $c_{n-1}$ words of length $n$ which satisfy the rules and whose first letter is $a$. Now if $l_1=b$, we must have that $l_2=a$, otherwise we would have two consecutive $b$'s. Once we place the $a$, there are no other affects that $l_1$ has on the rest of your word. This means that $l_3\dots l_n$ is a word of length $n-2$ satisfying your requirements. So there are $c_{n-2}$ words of length $n$ which have no consecutive $b$'s and whose first letter is a $b$. As each word can only start with $a$ or $b$, we have exhausted all possibilities. Thus we have the recurrence 
$$
c_n=c_{n-1}+c_{n-2},
$$
with the obvious initial conditions $c_1=2$ given by the words $a$ and $b$, and $c_2=3$ given by the words $aa$, $ab$, and $ba$. Can you go from here?
