# Difference between revisions of "2020 AMC 8 Problems/Problem 21"

## Problem 21

A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$-step paths are there from $P$ to $Q?$ (The figure shows a sample path.)

$[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("P", (5.5, 0.5)); label("Q", (6.5, 7.5)); [/asy]$

$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

## Solution

We count paths. Noticing that we can only go along white squares, to get to a white square we can only go from the two whites beneath it. Here is a diagram: $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("1", (5.5, .5)); label("1", (4.5, 1.5)); label("1", (6.5, 1.5)); label("1", (3.5, 2.5)); label("1", (7.5, 2.5)); label("2", (5.5, 2.5)); label("1", (2.5, 3.5)); label("3", (6.5, 3.5)); label("3", (4.5, 3.5)); label("4", (3.5, 4.5)); label("3", (7.5, 4.5)); label("6", (5.5, 4.5)); label("10", (4.5, 5.5)); label("9", (6.5, 5.5)); label("19", (5.5, 6.5)); label("9", (7.5, 6.5)); label("28", (6.5, 7.5)); [/asy]$ So the answer is $\boxed{\textbf{(A)}28}$ ~yofro (Diagram credits to franzliszt)

## Solution 2

Suppose we "extend" the chessboard indefinitely to the right:

$[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("P", (5.5,.5)); label("Q", (6.5,7.5)); label("X", (8.5,3.5)); label("Y", (8.5,5.5)); [/asy]$ The total number of paths from $P$ to $Q$ (including invalid paths which cross over the red line) is $\binom{7}{3} = 35$. We subtract the number of invalid paths that pass through $X$ or $Y$. The number of paths that pass through $X$ is $\binom{3}{0}\binom{4}{3} = 4$ and the number of paths that pass through $Y$ is $\binom{5}{1}\binom{2}{2} = 5$. However, we overcounted the invalid paths which pass through both $X$ and $Y$, of which there are 2 paths. Hence, the number of invalid paths is $4+5-2=7$ and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{\textbf{(A)} 28}$. -scrabbler94