# Difference between revisions of "2017 AMC 8 Problems/Problem 15"

## Problem

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path only allows moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

$[asy] fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray); fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray); label("8", (1, 0)); label("C", (2, 0)); label("8", (3, 0)); label("8", (0, 1)); label("C", (1, 1)); label("M", (2, 1)); label("C", (3, 1)); label("8", (4, 1)); label("C", (0, 2)); label("M", (1, 2)); label("A", (2, 2)); label("M", (3, 2)); label("C", (4, 2)); label("8", (0, 3)); label("C", (1, 3)); label("M", (2, 3)); label("C", (3, 3)); label("8", (4, 3)); label("8", (1, 4)); label("C", (2, 4)); label("8", (3, 4));[/asy]$

$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$

## Solution 1

Notice that the upper-most section contains a 3 by 3 square that looks like:

$[asy]label("8", (1, 2)); label("C", (2, 2)); label("8", (3, 2)); label("C", (1, 1)); label("M", (2, 1)); label("C", (3, 1)); label("M", (1, 0)); label("A", (2, 0)); label("M", (3, 0));[/asy]$

It has 6 paths in which you can spell out AMC8. You will find four identical copies of this square in the figure, so multiply ${6 \cdot 4}$ to get $\boxed{\textbf{(D)}\ 24}$ total paths.

## Solution 2

There are three different kinds of paths that are on this diagram. The first kind is when you directly count $A$, $M$, $C$ in a straight line. The second is when you count $A$, turn left or right to get $M$, then go up or down to count $8$ and $C$. The third is the one where you start with $A$, move up or down to count $M$, turn left or right to count $C$, then move straight again to get $8$.

There are 8 paths for each kind of path, making for $8 \cdot 3=\boxed{\textbf{(D)}\ 24}$ paths.

## Solution 3

Notice that the $A$ is adjacent to $4$ $M$s, each $M$ is adjacent to $3$ $C$s, and each $C$ is adjacent to $2$ $8$'s. Thus, the answer is $4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.$