Difference between revisions of "2024 AIME I Problems/Problem 6"

m
m (Image needed)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
Consider the paths of length <math>16</math> that follow the lines from the lower left corner to the upper right corner on an <math>8 \times 8</math> grid. Find the number of such paths that change direction exactly four times.
 
Consider the paths of length <math>16</math> that follow the lines from the lower left corner to the upper right corner on an <math>8 \times 8</math> grid. Find the number of such paths that change direction exactly four times.
 +
 +
{{image}}
  
 
==Solution 1==
 
==Solution 1==

Revision as of 20:47, 2 February 2024

Problem

Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8 \times 8$ grid. Find the number of such paths that change direction exactly four times.


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


Solution 1

We divide the path into eight “$R$” movements and eight “$U$” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$.


For $U$, we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$.

For $R$, we subtract $1$ from each section (to make the minimum stars of each section $0$) and we use Stars and Bars to get ${7 \choose 5}=21$.


Thus our answer is $7\cdot21\cdot2=\boxed{294}$.

~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 $7$ ways to choose when it will go up in the middle of the path and there are $\binom{7}{2}=21$ to choose the two places it will go right. Thus, there are $7\cdot21=147$ 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 $147\cdot2=\boxed{294}$

~alexanderruan

Solution 3

Notice that the $RURUR$ case and the $URURU$ 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 ${8+2-3 \choose 2}$ ways for the first part, and ${8+1-2 \choose 1}$ ways for the second part, by stars and bars.

The answer is $2\cdot {7 \choose 2} \cdot {7 \choose 1} = \boxed{294}$.

~northstar47

Feel free to edit this solution

See also

2024 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png