This is embarrasingly, the first problem in notes on introductory combinatorics by Polya and Tarjan. (Solved, but I havent looked).
Problem statement: Find the number of ways of spelling "abracadabra" always going from one letter to the adjacent.
$A$ $B \quad B$ $R \quad R \quad R $ $A\quad A \quad A \quad A $ $C\quad C\quad C\quad C\quad C$ $A\quad A \quad A \quad A \quad A\quad A$ $D\quad D\quad D\quad D\quad D$ $A\quad A \quad A \quad A $ $B \quad B \quad B $ $R \quad R$ $A$
I got a very improbable answer of $2^{25}$ so I tried a simpler case to understand it.
Starting at the northmost A there are two routes. At each of the two B's on the second row there are $2$ routes, so uptil this point
$A$ $B \quad B$ $R \quad R \quad R $
there should be $2^3$ ways to get the three letter word "ABR" but on manually counting the number of ways is just 4 (LL, LR, RR, RL where R=right/L=left). What is wrong with my approach? More precisely, where have I overcounted?
Edit: I understood my problem. I used the product rule instead of the sum rule. I think I will stop paying attention to these "rules" as they are hindering my problem solving anyway. (I have asked a previous questions on the subject)