Difference between revisions of "2014 AIME I Problems/Problem 11"

m (Solution (casework 2))
m (Solution (casework 2))
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==Solution (casework 2)==
 
==Solution (casework 2)==
Make a chart:
+
We split this into cases by making a chart. In each row, the entries <math>(\pm1)</math> before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position <math>(x,y)</math> satisfies <math>x=y</math>.
 +
<cmath>
 +
\begin{array}{ccccccccccccl}
 +
\multicolumn{5}{c}{R (+)\qquad L (-)}& |&\multicolumn{5}{c}{U (+)\qquad D (-)}\\
 +
+1&&  +1&&  +1&| &  +1&&  +1&&  +1\\
 +
+1&& +1&& -1& | & +1&& +1&& -1\\
 +
+1&& -1&&  -1& | & +1&& -1&& -1\\
 +
-1 && -1&&  -1& |  & -1&&  -1&& -1\\
 +
\\
 +
+1&& +1&& +1&& -1 &|& +1&& +1\\
 +
+1&& +1&& -1 && -1 &|& +1 && -1\\
 +
+1&& -1&&  -1 && -1 &|& -1 &&  -1 &&(\times 2 \text{ for symmetry by swapping }R-L\text{ and }U-D)\\
 +
\\
 +
+1&& +1 &&+1 &&-1&& -1& |& +1\\
 +
+1&& +1 &&-1&&  -1&& -1 &|& -1&& (\times 2\text{ symmetry})\\
 +
\\
 +
+1&& +1 &&+1&& -1&& -1 &&-1&| & (\times2 \text{ symmetry})\\
 +
\end{array}
 +
</cmath>
 +
Note that to account for the cases when <math>x=-y</math>, we can simply multiply the <math>U-D</math> steps by <math>-1</math>, so for example, the first row would become  <cmath>+1 \qquad+1\qquad +1 \ \ \ \ |\ \ \  -1\qquad -1\qquad -1.</cmath> Therefore, we must multiply the number of possibilities in each case by <math>2</math>, except for when <math>x=y=0</math>.
  
R (+) L (-) | U (+) D (-)
+
Now, we compute the number of possibilities for each case. In particular, we must compute the number of <math>RLUD</math> words, where <math>R</math> represents <math>+1</math> to the left of <math>|</math>, <math>L</math> represents <math>-1</math> to the left of <math>|</math>, <math>U</math> represents <math>+1</math> to the right of <math>|</math>, and <math>D</math> represents <math>-1</math> to the right of <math>|</math>. Using multinomials, we compute the following numbers of possibilities for each case.
  +++1 |  +++1
+
<cmath>{6\choose 3}\cdot 2+ \frac{6!}{2!2!}\cdot 2 + \frac{6!}{2!2!} \cdot 2 + {6\choose 3} \cdot 2 = 2(20 + 180 + 180 + 20) = 800</cmath>
+1 +1 -1  | +1 +1 -1
+
<cmath>\frac{6!}{3!2!}\cdot 2 + \frac{6!}{2!2!} + \frac{6!}{3!2!}\cdot 2 = 120 + 180 + 120 = 420\  (\times2\text{ for symmetry})</cmath>
+1 -1 -1  | +1 -1 -1
+
<cmath>\frac{6!}{3!2!} \cdot 2 + \frac{6!}{3!2!} \cdot 2 = 120 + 120 = 240\ (\times2\text{ for symmetry})</cmath>
-1  -1  -1  | -1  -1 -1
+
<cmath>{6\choose 3} = 20\ (\times 2\text{ for symmetry})</cmath>
  
+1 +1 +1 -1 | +1 +1
+
Thus, there are <math>800 + 840 + 480 + 40 = 2160</math> possibilities where <math>|x|=|y|</math>. Because there are <math>4^6</math> total possibilities, the probability is <math>\frac{2160}{4^6} = \frac{135}{256}</math>, so the answer is <math>\boxed{391}.</math>
+1 +1 -1  -1 | +1  -1
 
+1 -1  -1  -1 | -1  -1 (x2 symmetry by swapping R-L and U-D)
 
 
 
+1 +1 +1 -1 -1 | +1
 
+1 +1 -1  -1 -1 | -1 (x2 symmetry)
 
 
 
+1 +1 +1 -1 -1 -1 | (x2 symmetry)
 
 
 
Note we have to multiply by two to account for cases like +1 +1 +1 | -1 -1 -1.
 
 
 
Now, we use multinomials and do the arithmetic:
 
<cmath>{6\choose 3} * 2 + 6!/(2!2!) * 2 + 6!/(2!2!) * 2 + {6\choose 3} * 2 = 2(20 + 180 + 180 + 20) = 800</cmath>
 
<cmath>6!/(3!2!) * 2 + 6!/(2!2!) + 6!/(3!2!)) * 2 = 120 + 180 + 120 = 420 (*2)</cmath>
 
<cmath>6!/(3!2!) * 2 + 6!/(3!2!) * 2 = 120 + 120 = 240 (*2)</cmath>
 
<cmath>{6\choose 3} = 20 (*2)</cmath>
 
 
 
Thus, there are 800 + 840 + 480 + 40 = 2160 cases. Because there are <math>4^6</math> total cases, the answer is <math>\frac{2160}{4^6} = \frac{135}{256}</math>, so the answer is <math>\boxed{391}.</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2014|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:35, 13 March 2015

Problem 11

A token starts at the point $(0,0)$ of an $xy$-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution (casework)

We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$, or $4096$. There are 4 possible cases that land along the line $y = x$: $x,y = \pm 1; x,y = \pm 2; x,y = \pm 3;$ or $x = y = 0$. We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$, multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$.

  • Case 1: The token ends at $(3, 3)$. In order for the token to end up here, it must have had 3 right moves, and 3 up moves. In other words, the total number of ways to get here is the ways to rearrange the letters in the sequence $RRRUUU$, which is ${6\choose 3} = 20.$
  • Case 2: The token ends at $(2,2)$. In order for the token to end up here, it could have had 2 up moves, 3 right moves, and 1 left move; or 2 right moves, 3 up moves, and 1 down move. Thus, the total number of ways to get here is sum of the ways to rearrange the letters in the sequences $RRRLUU$ and $UUUDRR$, both of which are ${6\choose 1}{5\choose 2} = 60$, for a total of $120$ possibilities.
  • Case 3: The token ends at $(1,1)$. In order for the token to end up here, it could have had:
    • 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RUUUDD$ is ${6\choose 1}{5\choose 2} = 60.$
    • 1 up move, 3 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $URRRLL$ is ${6\choose 1}{5\choose 2} = 60.$
    • 2 right moves, 1 left move, 2 up moves, and 1 down move. In this case, the number of ways to rearrange the letters in the sequence $UUDRRL$ is ${6\choose 1}{5\choose 1}{4\choose 2} = 180.$

Thus, the total number of ways to end up at $(1,1)$ is $300$.

  • Case 4: The token ends at $(0,0)$. In order for the token to end up here, it could have had:
    • 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRRLLL$ is ${6\choose 3} = 20.$
    • 3 up moves and 3 down moves. In this case, the number of ways to rearrange the letters in the sequence $UUUDDD$ is ${6\choose 3} = 20.$
    • 1 right move, 1 left move, 2 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence $RLUUDD$ is ${6\choose 1}{5\choose 1}{4\choose 2} = 180.$
    • 1 up move, 1 down move, 2 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence $RRLLUD$ is ${6\choose 1}{5\choose 1}{4\choose 2} = 180.$

Thus, the total number of ways to end up at $(0,0)$ is $400$.

Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$. Thus, our probability is $\frac{2160}{4096}$. When this fraction is fully reduced, it is $\frac{135}{256}$, so our answer is $135 + 256 = \boxed{391}.$

Solution (casework 2)

We split this into cases by making a chart. In each row, the entries $(\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$. \[\begin{array}{ccccccccccccl} \multicolumn{5}{c}{R (+)\qquad L (-)}& |&\multicolumn{5}{c}{U (+)\qquad D (-)}\\ +1&&  +1&&  +1&| &  +1&&  +1&&  +1\\ +1&& +1&& -1& | & +1&& +1&& -1\\ +1&& -1&&  -1& | & +1&& -1&& -1\\ -1 && -1&&  -1& |  & -1&&  -1&& -1\\ \\ +1&& +1&& +1&& -1 &|& +1&& +1\\ +1&& +1&& -1 && -1 &|& +1 && -1\\ +1&& -1&&  -1 && -1 &|& -1 &&  -1 &&(\times 2 \text{ for symmetry by swapping }R-L\text{ and }U-D)\\ \\ +1&& +1 &&+1 &&-1&& -1& |& +1\\ +1&& +1 &&-1&&  -1&& -1 &|& -1&& (\times 2\text{ symmetry})\\ \\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\times2 \text{ symmetry})\\ \end{array}\] Note that to account for the cases when $x=-y$, we can simply multiply the $U-D$ steps by $-1$, so for example, the first row would become \[+1 \qquad+1\qquad +1 \ \ \ \ |\ \ \  -1\qquad -1\qquad -1.\] Therefore, we must multiply the number of possibilities in each case by $2$, except for when $x=y=0$.

Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$, $L$ represents $-1$ to the left of $|$, $U$ represents $+1$ to the right of $|$, and $D$ represents $-1$ to the right of $|$. Using multinomials, we compute the following numbers of possibilities for each case. \[{6\choose 3}\cdot  2+ \frac{6!}{2!2!}\cdot 2 + \frac{6!}{2!2!} \cdot 2 + {6\choose 3} \cdot 2 = 2(20 + 180 + 180 + 20) = 800\] \[\frac{6!}{3!2!}\cdot 2 + \frac{6!}{2!2!} + \frac{6!}{3!2!}\cdot 2 = 120 + 180 + 120 = 420\  (\times2\text{ for symmetry})\] \[\frac{6!}{3!2!} \cdot 2 + \frac{6!}{3!2!} \cdot 2 = 120 + 120 = 240\  (\times2\text{ for symmetry})\] \[{6\choose 3} = 20\ (\times 2\text{ for symmetry})\]

Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$. Because there are $4^6$ total possibilities, the probability is $\frac{2160}{4^6} = \frac{135}{256}$, so the answer is $\boxed{391}.$

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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