To plan a round trip from $A$ to $C$ and back, where the return route is not the reverse of the first part from $A$ to $C$, we proceed as follows. Since there are $14$ ways to go from $A$ to $C$ we must choose one of those ways. Once that has been done, exactly one of the $14$ routes has been ruled out for reversal to get back to $A$ from $C$, leaving us $13$ choices for the return trip. This gives $14\cdot 13=182$ possible round trips.
Note that we do not have to consider the specific types of these trips in terms of how they go through $B$ (or avoid $B$). One of the trips for the first leg of the journey from $A$ to $C$ being chosen, it is the only one excluded on the way back.
a, b)
Be $X$ a random variable that counts the repeating numbers from the right. Take into account that when a post is a Quads is not a Trips.
$\begin{align}
P(X=3)&=1\times\frac{1}{10}\times\frac{1}{10}\times\frac{9}{10}\\
&=0.9\%\\
P(X=4)&=1\times\frac{1}{10}\times\frac{1}{10}\times\frac{1}{10}\times\frac{9}{10}\\
&=0.09\%
\end{align}$
That means, you choose the rightests number, then have $\frac{1}{10}$ the next one is the same, and so on, until the fourth or the fifth, which you want it to be different.
c)
Take into account the following. If you have a Trips ($xxxxxx111$), the next one will come when you sum $111$, being $xxxxxx222$. There are two exceptions, though. First is when you get through the Quads. Where you'll have to sum $222$ from one to the next one, since one doesn't count. The other is when you are in $xxxxxx999$, where you get the next Trips on your next post, $xxxxxy000$.
Anyway, the mean for that is easier calculated using the probability we just found.
If we'd select random number of posts $t$, by mean we'd find $tP(X=3)$ Trips. Let's find for what number of posts, the expected number of Trips is $1$.
$\begin{align}
1&=tP(X=3)\\
&=0.0009t\\
t&=111.1111
\end{align}$
By mean, every $111.1111$ posts you'll have a Trips. For Quads it's equivalent.
$\begin{align}
1&=t'P(X=4)\\
&=0.00009t'\\
t&=1111.111
\end{align}$
By mean, every $1111.111$ posts you'll have a Quads.
Best Answer
There are $14$ ways to go there, and for each of those $14$ ways there are $14$ more ways to go back.
So, if you took route $1$ there, the number of routes back is $14$. If you took route $2$ there, the number of routes back is $14$ ($28$ total). Do this $14$ times because there are $14$ routes there, and the result is $14 \times 14$.