Difference between revisions of "2020 AMC 8 Problems/Problem 21"
Scrabbler94 (talk | contribs) (→Solution: add alternate solution without using recursion) |
Franzliszt (talk | contribs) (→Solution 2) |
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</asy> | </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 | 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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | The easiest way to solve this problem is just writing out the steps to get to each square as follows: | ||
+ | |||
+ | <asy> | ||
+ | size(200); | ||
+ | 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 thus <math>\textbf{(A) }28</math>. Interestingly, the final answer is also <math>\binom 82...</math> | ||
+ | |||
+ | -franzliszt | ||
==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 14:04, 18 November 2020
Problem 21
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
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: So the answer is ~yofro (Diagram credits to franzliszt)
Solution 2
Suppose we "extend" the chessboard indefinitely to the right:
The total number of paths from to (including invalid paths which cross over the red line) is . We subtract the number of invalid paths that pass through or . The number of paths that pass through is and the number of paths that pass through is . However, we overcounted the invalid paths which pass through both and , of which there are 2 paths. Hence, the number of invalid paths is and the number of valid paths from to is . -scrabbler94
Solution 3
The easiest way to solve this problem is just writing out the steps to get to each square as follows:
The answer is thus . Interestingly, the final answer is also
-franzliszt
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.