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

## Problem 15

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 allows only 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

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 $4\cdot 3\cdot 2 = \boxed{\textbf{(D)}\ 24}.$

~nukelauncher

## Alternate Solution

[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] 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 straight to count M and C. The third is the one where you start with A, move forward 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 $8 \cdot 3=\boxed{24}$ paths.

-coolmath34