Difference between revisions of "2014 AIME I Problems/Problem 11"
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Adding these cases together, we get that the total number of ways to end up on <math>y = x</math> is <math>4(20 + 120 + 300) + 400 = 2160</math>. Thus, our probability is <math>\frac{2160}{4096}</math>. When this fraction is fully reduced, it is <math>\frac{135}{256}</math>, so our answer is <math>135 + 256 = \boxed{391}.</math> | Adding these cases together, we get that the total number of ways to end up on <math>y = x</math> is <math>4(20 + 120 + 300) + 400 = 2160</math>. Thus, our probability is <math>\frac{2160}{4096}</math>. When this fraction is fully reduced, it is <math>\frac{135}{256}</math>, so our answer is <math>135 + 256 = \boxed{391}.</math> | ||
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Can other people see this? | Can other people see this? | ||
+ | -syuan01 | ||
==Solution (casework 2)== | ==Solution (casework 2)== |
Revision as of 16:54, 20 February 2019
Problem 11
A token starts at the point of an
-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
is
, where
and
are relatively prime positive integers. Find
.
Solution (casework)
We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is , or
. There are 4 possible cases that land along the line
:
or
. We will count the number of ways to end up at
and
, multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at
.
- Case 1: The token ends at
. 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
, which is
- Case 2: The token ends at
. 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
and
, both of which are
, for a total of
possibilities.
- Case 3: The token ends at
. 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
is
- 1 up move, 3 right moves, and 2 left moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 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
is
- 1 right move, 3 up moves, and 2 down moves. In this case, the number of ways to rearrange the letters in the sequence
Thus, the total number of ways to end up at is
.
- Case 4: The token ends at
. 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
is
- 3 up moves and 3 down moves. In this case, the number of ways to rearrange the letters in the sequence
is
- 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
is
- 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
is
- 3 right moves and 3 left moves. In this case, the number of ways to rearrange the letters in the sequence
Thus, the total number of ways to end up at is
.
Adding these cases together, we get that the total number of ways to end up on is
. Thus, our probability is
. When this fraction is fully reduced, it is
, so our answer is
Can other people see this? -syuan01
Solution (casework 2)
We split this into cases by making a chart. In each row, the entries 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
satisfies
.
Note that to account for the cases when
, we can simply multiply the
steps by
, so for example, the first row would become
Therefore, we must multiply the number of possibilities in each case by
, except for when
.
Now, we compute the number of possibilities for each case. In particular, we must compute the number of words, where
represents
to the left of
,
represents
to the left of
,
represents
to the right of
, and
represents
to the right of
. Using multinomials, we compute the following numbers of possibilities for each case.
Thus, there are possibilities where
. Because there are
total possibilities, the probability is
, so the answer is
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.