# Number of ways of spelling Abracadabra in this grid

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)

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I think your problem is that you're treating picking the "A" as one of two choices, when there's really nothing you can do about it. Two routes to a B and two from each B to an R gives $2*2 = 4$ ways. – El'endia Starman May 23 '11 at 3:33
Please don't add "SOLVED" to the subject. The main site notes that an answer has been accepted, that is more than sufficient. – Arturo Magidin May 23 '11 at 4:24
@El'enida Thanks. I think my problem is formulating the problem in words and then try to solve it. I guess combinatoric problems are best solved intuitively. – kuch nahi May 23 '11 at 4:25
@Arturo Ok. And thanks for answering. – kuch nahi May 23 '11 at 4:26

This is essentially the same as this problem. Start by rewriting your array as a square, with the top now at the bottom left, and the bottom now at the top right: $$\begin{array}{cccccc} A & D & A & B & R & A\\ C & A & D & A & B & R\\ A & C & A & D & A & B\\ R & A & C & A & D & A\\ B & R & A & C & A & D\\ A & B & R & A & C & A \end{array}$$ You want to get from the bottom left to the top right, using only moves right or up. A simple thing is to note that there is only one way to get to the bottom left, and at any point, the number of ways to get to a point is the sum of the number of ways to get to the point directly below and the number of ways to get to the point directly left. This gives: $$\begin{array}{cccccc} 1 & 6 & 21 & 56 & 126 & 252\\ 1 & 5 & 15 & 35 & 70 & 126\\ 1 & 4 & 10 & 20 & 35 & 56\\ 1 & 3 & 6 & 10 & 15 & 21\\ 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 1 & 1 & 1 & 1 & 1 \end{array}$$ giving $252$ ways.
Alternatively, in the end, you must make five moves right and five moves up. Your only decisions to make is when you make the moves up. You have six locations (before all the moves right, between any two moves right, and after all moves right), and you have to choose them, allowing repetitions and without caring about the order. That is, you want to count combinations with repetitions, selecting six from seven possibililties. The way to make $r$ choices from $n$ possibilities, allowing repetitions and where order does not matter, is $\binom{n+r-1}{r}$, so in this case we have $$\binom{6+5-1}{5} = \binom{10}{5} = \frac{10!}{5!5!} = 252.$$
It was quite clever to have reformulated it this way. Inspired by the class of problems you've exposed this to, another restatement could be: number of non-negative solutions of $x_1+\cdots +x_6=5$ (based on the first grid you have drawn) – kuch nahi May 23 '11 at 4:03
+1 though I am not sure why you need to rewrite the pattern: in the original you need 5 moves south-west and 5 moves south-east, and can choose these in ${5+5 \choose 5}=252$ different orders. – Henry May 23 '11 at 6:39