Difference between revisions of "2020 AMC 8 Problems/Problem 21"
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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>. | 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== | ==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 18:36, 23 November 2020
Problem
A game board consists of squares that alternate in color between black and white. The figure below shows square
in the bottom row and square
in the top row. A marker is placed at
A step consists of moving the marker onto one of the adjoining white squares in the row above. How many
-step paths are there from
to
(The figure shows a sample path.)
Solution 1
Notice that, in order to step onto any particular white square, the marker must have come from one of the or
white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from
to that square is the sum of the numbers of ways to move from
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
to that square, as already stated).
The answer is therefore
.
Solution 2
Suppose we "extend" the chessboard infinitely with additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.
The total number of paths from
to
, including invalid paths which cross over the red line, is then the number of paths which make
steps up-and-right and
steps up-and-left, which is
. We need to subtract the number of invalid paths, i.e. the number of paths that pass through
or
. To get to
, the marker has to make
up-and-right steps, after which it can proceed to
with
steps up-and-left and
step up-and-right. Thus, the number of paths from
to
that pass through
is
. Similarly, the number of paths that pass through
is
. However, we have now double-counted the invalid paths which pass through both
and
; from the diagram, it is clear that there are only
of these (as the marker can get from
to
by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is
, and the number of valid paths from
to
is
.
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.