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

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So the answer is <math>\boxed{\textbf{(A)}28}</math>
 
So the answer is <math>\boxed{\textbf{(A)}28}</math>
 
~yofro (Diagram credits to franzliszt)
 
~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 <math>P</math> to <math>Q</math> (including invalid paths which cross over the red line) is <math>\binom{7}{3} = 35</math>. We subtract the number of invalid paths that pass through <math>X</math> or <math>Y</math>. The number of paths that pass through <math>X</math> is <math>\binom{3}{0}\binom{4}{3} = 4</math> and the number of paths that pass through <math>Y</math> is <math>\binom{5}{1}\binom{2}{2} = 5</math>. However, we overcounted the invalid paths which pass through both <math>X</math> and <math>Y</math>, of which there are 2 paths. 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>. -scrabbler94
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=20|num-a=22}}
 
{{AMC8 box|year=2020|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:27, 18 November 2020

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

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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