Difference between revisions of "2024 AIME I Problems/Problem 6"
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Feel free to edit this solution | Feel free to edit this solution | ||
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+ | ==Solution 4== | ||
+ | Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle <math>7</math>x<math>7</math> grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point <math>(3,2)</math> and <math>(5,3)</math>, you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have <math>7</math> points on each column to choose from and starting from left to right, we have <math>6,5,4,3,2,1</math> points on that row to choose from. This gives us <math>7(6)+7(5)+7(4)+7(3)+7(2)+7(1)</math> which simplifies to <math>7\cdot21</math>. The vertical case is symmetrical so we have <math>7\cdot21\cdot2 = \boxed{294}</math> | ||
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+ | ~KEVIN_LIU | ||
==See also== | ==See also== |
Revision as of 21:51, 3 February 2024
Problem
Consider the paths of length that follow the lines from the lower left corner to the upper right corner on an
grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
Solution 1
We divide the path into eight “” movements and eight “
” movements. Five sections of alternative
or
are necessary in order to make four “turns.” We use the first case and multiply by
.
For , we have seven ordered pairs of positive integers
such that
.
For , we subtract
from each section (to make the minimum stars of each section
) and we use Stars and Bars to get
.
Thus our answer is .
~eevee9406
Solution 2
The path can either start by going up or start by going right. Suppose it starts by going up. After a while, it will turn to the right. Then, it will go up. After that, it will go right again. There are ways to choose when it will go up in the middle of the path and there are
to choose the two places it will go right. Thus, there are
ways to create a path that starts by going up. By symmetry, this is the same as the number of paths that start by going right, so the answer is
~alexanderruan
Solution 3
Notice that the case and the
case is symmetrical. WLOG, let's consider the RURUR case.
Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of balls.
There are ways for the first part, and
ways for the second part, by stars and bars.
The answer is .
~northstar47
Feel free to edit this solution
Solution 4
Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle x
grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point
and
, you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have
points on each column to choose from and starting from left to right, we have
points on that row to choose from. This gives us
which simplifies to
. The vertical case is symmetrical so we have
~KEVIN_LIU
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.