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

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==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, as in the examples shown below.
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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.
  
 
==Solution==
 
==Solution==
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For <math>U</math>, we have seven ordered pairs of positive integers <math>(a,b)</math> such that <math>a+b=8</math>.
 
For <math>U</math>, we have seven ordered pairs of positive integers <math>(a,b)</math> such that <math>a+b=8</math>.
  
For <math>R</math>, we subtract <math>1</math> from each section (as the minimum is <math>1</math>) and we use Stars and Bars to get <math>(7 \choose 5)=21</math>.
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For <math>R</math>, we subtract <math>1</math> from each section (to make the minimum stars of each section <math>0</math>) and we use Stars and Bars to get <math>{7 \choose 5}=21</math>.
  
  
Thus our answer is <math>7*21*2=\boxed{294}</math>.
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Thus our answer is <math>7\cdot21\cdot2=\boxed{294}</math>.
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==See also==
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{{AIME box|year=2024|n=I|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 18:17, 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.

Solution

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 $7\cdot21\cdot2=\boxed{294}$.


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

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