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

Revision as of 17:36, 23 November 2020

Problem

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 1

Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[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]$ The answer is therefore $\boxed{\textbf{(A) }28}$ .

Solution 2

Suppose we "extend" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.

$[asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("$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 then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\binom{4+3}{3} = \binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$ . To get to $X$ , the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through $Y$ is $\binom{4+1}{1}\cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both $X$ and $Y$ ; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$ , and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{\textbf{(A) }28}$ .

Video Solution

https://youtu.be/YR3oghbziBA

~savannahsolver

@@ Line 59: / Line 59: @@
 </asy>
 The total number of paths from <math>P</math> to <math>Q</math>, including invalid paths which cross over the red line, is then the number of paths which make <math>4</math> steps up-and-right and <math>3</math> steps up-and-left, which is <math>\binom{4+3}{3} = \binom{7}{3} = 35</math>. We need to subtract the number of invalid paths, i.e. the number of paths that pass through <math>X</math> or <math>Y</math>. To get to <math>X</math>, the marker has to make <math>3</math> up-and-right steps, after which it can proceed to <math>Q</math> with <math>3</math> steps up-and-left and <math>1</math> step up-and-right. Thus, the number of paths from <math>P</math> to <math>Q</math> that pass through <math>X</math> is <math>1 \cdot \binom{3+1}{3} = 4</math>. Similarly, the number of paths that pass through <math>Y</math> is <math>\binom{4+1}{1}\cdot 1 = 5</math>. However, we have now double-counted the invalid paths which pass through both <math>X</math> and <math>Y</math>; from the diagram, it is clear that there are only <math>2</math> of these (as the marker can get from <math>X</math> to <math>Y</math> by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is <math>4+5-2=7</math>, and the number of valid paths from <math>P</math> to <math>Q</math> is <math>35-7 = \boxed{\textbf{(A) }28}</math>.
+==Video Solution==
+https://youtu.be/YR3oghbziBA
+~savannahsolver
 ==See also==
 {{AMC8 box|year=2020|num-b=20|num-a=22}}
 {{MAA Notice}}

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